Here’s the set of ordered pairs $A\times\wp(A)$, with $A$ listed across the bottom and $\wp(A)$ along the lefthand edge:
$$\begin{array}{r|ll}
\{\varnothing,\{\varnothing\}\}&\langle\varnothing,\{\varnothing,\{\varnothing\}\}\rangle&\langle\{\varnothing\},\{\varnothing,\{\varnothing\}\}\rangle\\
\{\{\varnothing\}\}&\langle\varnothing,\{\{\varnothing\}\}\rangle&\langle\{\varnothing\},\{\{\varnothing\}\}\rangle\\
\{\varnothing\}&\langle\varnothing,\{\varnothing\}\rangle&\langle\{\varnothing\},\{\varnothing\}\rangle\\
\varnothing&\langle\varnothing,\varnothing\rangle&\langle\{\varnothing\},\varnothing\rangle\\ \hline
&\varnothing&\{\varnothing\}
\end{array}\tag{1}$$
Here’s the same table of $B\times\wp(B)$, where $B=\{0,1\}$:
$$\begin{array}{r|ll}
\{0,1\}&\langle0,\{0,1\}\rangle&\langle1,\{0,1\}\rangle\\
\{1\}&\langle0,\{1\}\rangle&\langle1,\{1\}\rangle\\
\{0\}&\langle0,\{0\}\rangle&\langle1,\{0\}\rangle\\
\varnothing&\langle0,\varnothing\rangle&\langle1,\varnothing\rangle\\ \hline
&0&1
\end{array}\tag{2}$$
As you can see, they have exactly the same structure: only the labels are different. If you go through $(1)$ replacing $\varnothing$ by $0$ and $\{\varnothing\}$ by $1$, you get exactly $(2)$.
And here, to make the structure even clearer, is $B\times C$, where $C=\{0,1,2,3\}$
$$\begin{array}{r|ll}
3&\langle0,3\rangle&\langle1,3\rangle\\
2&\langle0,2\rangle&\langle1,2\rangle\\
1&\langle0,1\rangle&\langle1,1\rangle\\
0&\langle0,0\rangle&\langle1,0\rangle\\ \hline
&0&1
\end{array}\tag{3}$$
If in $(2)$ you replace every second coordinate $\varnothing$ by $0$, every second coordinate $\{0\}$ by $1$, every second coordinate $\{1\}$ by $2$, and every second coordinate $\{0,1\}$ by $3$, you get exactly $(3)$. The three Cartesian products $A\times\wp(A)$, $B\times\wp(B)$, and $B\times C$ have the same basic structure. This structure is easiest to see in $(3)$, but that’s only because the notation is less cluttered.
Added: I think that the problem that you’re having isn’t really with the ordered pairs themselves, but rather with the objects appearing in them. In practical terms an ordered pair is just a list of two things arranged so that there is a first thing and a second thing (which may be the same thing listed twice), and we can tell which is which. As far as the ordered pair aspect is concerned, there’s no difference between $\langle\varnothing,\{\{\varnothing\}\}\rangle$ and $\langle 3,7\rangle$: both are ordered pairs whose first and second components are unequal. Each becomes a different ordered pair if you reverse the order of its components: $\langle\{\{\varnothing\}\},\varnothing\rangle\ne\langle\varnothing,\{\{\varnothing\}\}\rangle$, and $\langle 7,3\rangle\ne\langle3,7\rangle$.
To put it a little differently, the meaning of $$\langle\text{thing}_1,\text{thing}_2\rangle$$ is simply list of two things, the first one being $\text{thing}_1$ and the second $\text{thing}_2$. The identities of $\text{thing}_1$ and $\text{thing}_2$ have no bearing on this meaning. We might say that they’re buried one layer deeper: after identifying $\langle\text{thing}_1,\text{thing}_2\rangle$ as an ordered pair of entities, we can worry about just what those entities are. If $\text{thing}_1=3$ and $\text{thing}_2=7$, and the context is plotting a point in the Cartesian plane, we hardly notice this step, because we’re very familiar with $3$ and $7$ as real numbers. If $\text{thing}_1=3$ and $\text{thing}_2=7$, and the context is a first serious course in set theory, we might have to think a bit harder about this step, because in that context it’s likely that $3=\{0,1,2\}$ and $7=\{0,1,2,3,4,5,6\}$, so that
$$\langle3,7\rangle=\langle\{0,1,2\},\{0,1,2,3,4,5,6\}\rangle\;.$$
But the added complication has nothing to do with the ordered pair structure: it’s all in the two objects that are the components of the ordered pair.
The same is true if $\text{thing}_1=\varnothing$ and $\text{thing}_2=\{\{\varnothing\}\}$: $\langle\varnothing,\{\{\varnothing\}\}\rangle$ is just a list whose first element is $\varnothing$, and whose second element is $\{\{\varnothing\}\}$. If you don’t feel that you have an intuitive grip on this, the problem is unlikely to be in the ordered pair structure itself, in the notion of a first thing and a second thing; it’s much likelier to result from the relative unfamiliarity of the things themselves. That’s a problem that to a large extent will solve itself over time, provided that you keep working with the concepts. In the short term it may help to make a conscious effort to think about ordered pairs in ‘layers’:
Okay, this is an ordered pair. Its first component is $\text{thing}_1$, and its second component is $\text{thing}_2$. It’s an element of the function $f$, so I know that $f(\text{thing}_1)=\text{thing}_2$. Now what are these things? Well, $\text{thing}_1=\varnothing$ and $\text{thing}_2=\{\{\varnothing\}\}$, and therefore $f(\varnothing)=\{\{\varnothing\}\}$. Okay: the function $f$ assigns to the empty set the set $\{\{\varnothing\}\}$. Do I really need to know more than that right now?
In this example the ordered pair is in a middle layer, with the function $f$ ‘above’ it and the objects $\text{thing}_1$ and $\text{thing}_2$ ‘below’ it. It’s part of the internal structure of $f$, and $\text{thing}_1$ and $\text{thing}_2$ are part of its internal structure. They in turn may have internal structure; in this case $\varnothing$ has none, but $\{\{\varnothing\}\}$ clearly does have some. The details of that internal structure may or may not matter. If they do, I’ll have to think about them: $\{\{\varnothing\}\}$ is the set whose only element is $\{\varnothing\}$, which in turn is the set whose only element is $\varnothing$. If not, I can just treat $\{\{\varnothing\}\}$ as a fancy label for some set whose precise nature isn’t important, at least at the moment. If it becomes important later, I can worry about it then.
Actually, this advice applies to all mathematical notation. You don’t have to grasp a complicated expression all at once: it’s fine to build up an understanding of it a piece at a time. An equation, for instance, by definition has the form $A=B$. We see that, and we immediately know the type of expression with which we’re dealing. Now what are $A$ and $B$? And we go on from there, making sense of $A$ and $B$.
Best Answer
Whatever it is we define "$(a,b)$" to be as a set, what we really want is the following "defining property":
There are many ways to achieve this, but this is what we really want to achieve; once we achieve this via some definition, we want to avoid using the actual "guts" of the definition and stick exclusively to that defining property. (Similar to the points made in this answer and comment about how to represent the real numbers as sets).
One way to achieve this "defining property" is via the Kuratowski definition, by defining $$(a,b) = \Bigl\{ \{a\},\{a,b\}\Bigr\}.$$ We can prove that this is a set, and that this set has the property we want.
There are other ways of achieving the same result; for example, Wiener proposed $$(a,b) = \Bigl\{ \bigl\{ \{a\},\emptyset\bigr\}, \bigl\{\{b\}\bigr\}\Bigr\},$$ which also has the "defining property".
The problem with your proposal is that it does not have the defining property we want for ordered pairs: for example, $\emptyset\neq\{\emptyset\}$, so we want $(\emptyset,\{\emptyset\}) \neq (\{\emptyset\},\emptyset)$. But in your proposal, we have: $$\begin{align*} (\emptyset,\{\emptyset\}) &= \Bigl\{ \{\emptyset\}, \bigl\{\{\emptyset\}\bigr\}, \bigl\{ \emptyset, \{\emptyset\}\bigr\}\Bigr\},\\ (\{\emptyset\},\emptyset) &= \Bigl\{ \bigl\{ \{\emptyset\}\bigr\}, \{\emptyset\}, \bigl\{ \{\emptyset\},\emptyset\bigr\}\Bigr\}; \end{align*}$$ so that $(\emptyset,\{\emptyset\}) = (\{\emptyset\},\emptyset)$.
So the proposal, while a perfectly fine definition of a set, does not achieve the ultimate purpose of defining the ordered pair, and so it should not be the definition of "ordered pair".
Proof that the Kuratowski definition has the "defining property".
If $a=c$ and $b=d$, then $$(a,b) = \Bigl\{ \{a\}, \{a,b\}\Bigr\} = \Bigl\{ \{c\},\{c,d\}\Bigr\} = (c,d).$$ Assume conversely that $(a,b)=(c,d)$. Then $\bigcap(a,b) = \bigcap(c,d)$. Since $$\bigcap(a,b) = \bigcap\Bigl\{ \{a\},\{a,b\}\Bigr\} = \{a\}\cap\{a,b\} = \{a\}$$ and $$\bigcap(c,d) = \bigcap\Bigr\{ \{c\}, \{c,d\}\Bigr\} = \{c\}\cap\{c,d\} = \{c\},$$ we conclude that $a=c$.
If $b=a$, then $(a,b) = \{\{a\}\} = (c,d) = \{\{c\},\{c,d\}\}$. Therefore, $\{c,d\} \in\{\;\{a\}\;\}$, so $d\in\{a\}$, hence $d=a=b$ and we conclude $d=b$, as desired. Symmetrically, if $c=d$, then $\{a,b\}\in(a,b)=(c,d) = \{\;\{c\}\;\}$, so $\{a,b\}=\{c\}$, hence $b=c=d$ and we again conclude $d=b$ as desired.
If $b\neq a$ and $c\neq d$, then $\bigcup(a,b)-\bigcap(a,b) = \bigcup(c,d)-\bigcap(c,d)$. Since $$\bigcup(a,b)-\bigcap(a,b) = \bigcup\Bigl\{\{a\},\{a,b\}\Bigr\} - \{a\} = \Bigl( \{a\}\cup \{a,b\}\Bigr)-\{a\} = \{a,b\}-\{a\} = \{b\}$$ and $$\bigcup(c,d)-\bigcap(c,d) = \bigcup\Bigl\{\{c\},\{c,d\}\Bigr\} - \{c\} = \Bigl(\{c\}\cup \{c,d\}\Bigr)-\{c\} = \{c,d\}-\{c\} = \{d\}$$ (where we've used that $a\neq b$ to conclude that $\{a,b\}-\{a\}=\{b\}$ and we've used $c\neq d$ to conclude $\{c,d\}-\{c\}=\{d\}$), then we have $\{b\}=\{d\}$, hence $b=d$, again as desired.
Thus, if $(a,b)=(c,d)$, then $a=c$ and $b=d$.
Addressing the comments added to the question.
This definition is part of a way to try to define a lot of the things that we use in mathematics on the basis of an axiomatic theory; in this case, we start with Axiomatic Set Theory, where the only notions we have (if we are working in Zermelo-Fraenkel Set Theory) are "set" and "is and element of", together with the axioms that tells us properties of sets and things we can do with sets.
We want to have something that works like what we know as "the ordered pair"; but all we have to work with are sets. So we need to find a way of constructing a set that has the properties we want for the ordered pair.
For a metaphor: the ordered pair is like a car; we know how to drive. But in order to actually have a car, there needs to be an engine and gasoline, and the engine has to work. We are trying to construct that engine so that we can later drive it.
So this is not notation, this is a definition of what the ordered pair is in set theory. We are defining an object, which we call "$(a,b)$", to be the given set. It's not merely how we are writing the ordered pair, is what the ordered pair is if you are interested in actually seeing the engine of the car working. We know what we want "ordered pair" to behave like, but we have to actually construct an object that behaves that way. This is a way of defining an object that does behave that way.
There aren't "two notations" here. We define "the ordered pair with first component $a$ and second component $b$" to be the set $$\bigl\{ \{a\}, \{a,b\}\bigr\},$$ (which one can prove is indeed a set using the Axioms of Set Theory, if $a$ and $b$ are already in the theory).
Then we prove that "the ordered pair with first component $a$ and second component $b$" is equal to "the ordered pair with first component $c$ and second component $d$" if and only if $a=c$ and $b=d$.
Then we abbreviate "the ordered pair with first component $a$ and second component $b$" by writing "$(a,b)$" (or sometimes "$\langle a,b\rangle$").
"$(a,b)"$ is notation. The other side is the definition of this set.
The definition is the way it is because it works; that's really all we care about. In fact, we forget about the definition pretty much as soon as we can, and simply use the $(a,b)$ and the "defining property." We can do that, because we know that "under the hood" there actually is an engine that does what we need it to do, even if we don't see it working while we are driving the car.
So, there is only one bit of notation, and it's "$(a,b)$". The other side is the definition of what that notation actually is.
The "set notation" doesn't "say" the ordered pair is what you think it is. What we are doing is defining what an ordered pair is, in a theory where the only thing we have are sets and the axioms of set theory. Because sets don't respect order, we cannot rely on simply how we write something; in order to be able to define an ordered pair we need to give a purely set-theoretic definition that actually achieves the purpose we want. Kuratowski's definition of an ordered pair $(a,b)$ to be the set given by $\bigl\{\{a\},\{a,b\}\bigr\}$ achieves this objective, in that the defined object has precisely the property we want an "ordered-pair-whatever-it-may-actually-be" to have. Since this set has that property, we define that set to be what the ordered pair "really is". But we don't actually care about what an ordered pair "really is", we just care about its desired "defining property".
In order for your car to work, there has to be an engine somewhere; but once there is an engine and your car works, you don't need to see the engine working in order to drive the car. The same with the ordered pair: for us to have an "ordered pair" in set theory, we need to be able to construct it somehow using sets. Once we have managed to do that, we don't need to see the actual set, we can just use the fact that there is a set that achieves our desired goal.
Yes, two sets are equal if and only if they have the same elements. So $\bigl\{\{a\},\{a,b\}\bigr\} = \bigl\{ \{a,b\}, \{a\}\bigr\}$. This does not matter. What matters is that $\bigl\{ \{a\}, \{a,b\}\bigr\} = \bigl\{ \{c\},\{c,d\}\bigr\}$ if and only if $a=c$ and $b=d$, because that's what we are going for. The definition of ordered pair by Kuratowski is specifically designed so that the end result "encodes" an order and distinguishes between the "first component" and the "second component" of $(a,b)$. The definition actually achieves this, as I proved above.
There is no problem with "duplicate elements". It's just that the set $\bigl\{\{a\},\{a\}\bigr\}$ is equal to the set $\bigl\{\{a\}\bigr\}$ by the Axiom of Extension, which says that two sets $A$ and $B$ are equal if and only if for every $x$, $x\in A\leftrightarrow x\in B$. The ordered pair $(a,a)$, as a set, is a set which can be written as $$\bigl\{ \{a\},\{a,a\}\bigr\} \text{ or as }\bigl\{\{a\},\{a\}\bigr\}\text{ or as }\bigl\{ \{a\}\bigr\}.$$ There is no problem with this, because that set has the property that $(c,d)$ is equal to $\bigl\{ \{a\}\bigr\}$ if and only if $c=d=a$, which is exactly what we want.
Again: the whole point of this definition is only that it satisfies the property
Once we have this property, we abbreviate the set $\bigl\{\{a\},\{a,b\}\bigr\}$ as $(a,b)$, and simply use the property listed above.