Suppose $A \neq B$. Without loss of generality, there exists an $x \in A$ such that $x \notin B$. Then $\{x\} \in \mathscr{P}(A)$ whereas $\{x\} \notin \mathscr{P}(B)$. Thus $\mathscr{P}(A) \neq \mathscr{P}(B)$.
Conversely, if $\mathscr{P}(A) = \mathscr{P}(B)$, then all their singletons are the same. Thus $A = B$.
$A = B$ if and only if $\mathscr{P}(A) = \mathscr{P}(B)$.
It is not true for most sets $A$, because the elements of $A\times A$ are ordered pairs, and an ordered pair of elements from $A$ is not a usually set of elements of $A$ (which it would have to be in order to be a member of $\mathcal P(A)$.
In axiomatic set theory, where sets are the only things that exist, an ordered pair must be represented by a particular set -- the most common convention is to consider $(a,b)$ to "really" be an abbreviation for $\{\{a\},\{a,b\}\}$.
When this convention is used, there are a few particular cases where $A\times A$ is indeed a subset of $\mathcal P(A)$.
One example of this is $A=\varnothing$, in which case $A\times A=\varnothing$ which is a subset of anything. A different example is $A=H_\omega$, the set of hereditarily finite sets (which means basically the sets that can be written in finite space with only the symbols {, }, and ,, such as "$\{\{\},\{\{\}\},\{\{\},\{\{\}\}\}\}$").
But in both of those cases $A\times A\subseteq \mathcal P(A)$ is more of an accident than something that tells us something interesting and useful about the sets in question.
Best Answer
The first statement, $\mathcal{P}(A \times B) = \mathcal{P}(A) \times \mathcal{P}(B)$, is false. To disprove this, all you have to do is consider a special case like $A = \{1\}, B = \{1\}$. What are the two sets in this case? Or, count the sizes of the two sets.
The second statement is true. To prove the equality of sets $X = Y$, a very common strategy is to first show $X \subseteq Y$, and then show $Y \subseteq X$. To show $X \subseteq Y$, you start with an arbitrary $x \in X$ and you prove that $x \in Y$.
So we want to show the equality $\mathcal{P}(A \cap B) = \mathcal{P}(A) \cap \mathcal{P}(B)$. Suppose $x \in \mathcal{P}(A \cap B)$. Then $x \subseteq A \cap B$. It follows since $A \cap B \subseteq A$ and $A \cap B \subseteq B$ that $x \subseteq A$ and $x \subseteq B$. So $x \in \mathcal{P}(A)$ and $x \in \mathcal{P}(B)$. So $x \in \mathcal{P}(A) \cap \mathcal{P}(B)$.
That proves $\mathcal{P}(A \cap B) \subseteq \mathcal{P}(A) \cap \mathcal{P}(B)$. Can you do the other direction?