Your approach is sound. I think you can choose a better injection from $P(\mathbb{N})$ into equivalence relations. Let us say $0 \in \Bbb N$ and we will inject $P(\Bbb N \setminus\{0\})$ into the equivalence relations on $\Bbb N$. I would suggest you take a subset of $\Bbb N \setminus\{0\}$ to the equivalence relation that groups all members of the subset and $0$ into an equivalence class and leaves all the rest of $\Bbb N$ under the identity.
If we don't do the trick with $0$, all singleton subsets will be mapped to the identity relation and you don't have an injection.
The deficulty is understanding concepts and using language. If you can do that this is trivial (almost literally).
Let me clearify with a few examples.
Ex: 1) Let $A=\{dog, cat, mouse\}$
Ex: 2) Let $A =\mathbb R$.
Then $B = \{f:\{1\}\to A\}= \{K\subset \{1\}\times A|$ for each $x \in \{1\}$ there is exactly one $(x,y) \in K\}=$
$\{K\subset \{1\}\times A|$ there is exactly one $(1,y) \in K\}=$
$\{K\subset \{1\}\times A| K = \{(1,y)\}$ for some $y \in A\}=$
$\color{blue}{\big\{}\{(1,y)\}| y\in A\color{blue}{\big\}}$.
In example 1: then $B = \color{blue}{\big\{}\{(1,dog)\}, \{(1,cat)\}, \{(1,mouse)\}\color{blue}{\big\}}$.
In example 2: then $B = \color{blue}{\big\{}\small\{(1, y)\small\}|y\in \mathbb R\color{blue}{\big\}}$
... Now try to do this on your own without reading further ...
Now it should be intuitively obvious that for every $a \in A$ there is exactly one function $f: \{1\}\to A$ so that $f(1) = a$.
And that's it. That's your bijection:
...... try to formally define the bijection, $j: B \to A$, before reading further ......
Let $j: B \to A$ via for any $f \in B$ we set $j(f) = f(1)$.
... Now try to prove that that is an injection without reading further...
To formally prove $j$ is a bijection.
Surjective: For each $a\in A$ then if we define $f:\{1\} \to A$ as $f(1) =a$ then $f \in B$ and $j(f) = a$. So $j$ is surjective.
Injective: If $j(f) = j(g)= y$ for some $y \in A$ then $f(1) = y$ and $g(1) = y$. but then (as $1$ is the only element of $\{1\}$) for all $x \in \{1\}$ then $f(x) = g(x)$. So $f = g$. So $j$ is one to one.
Best Answer
Your explanation is absolutely correct. For the first one, you could subtract $1$ because people usually define a relation to be a nonempty subset of $A$x$B$. Empty relations are not that good.:p