Don't understand how $B \in \{\mathscr P(A) | A \in \mathscr F\} \equiv \exists A \in \mathscr F(B = \mathscr P(A))$ While I have seen and can understand a similar conversion in a different context, I can't seem to wrap my head around how this change works:
$B \in \{\mathscr P(A) | A \in \mathscr F\} \equiv \exists A \in \mathscr F(B = \mathscr P(A))$
In a simpler context, I can see that 
$x \in \{n^2 | n \in \mathbb N\} \equiv \exists n \in \mathbb N(x = n^2)$
Because the former would be described as "x is a member of the set of all n-squared such that n is a member of the set of natural numbers" while the latter would be "There exists a number n in the set of natural numbers for which x is equal to n-squared." Obviously if x is a member of the set of square numbers, it makes sense that there must be a number in the given universe of discourse that can be squared to get x. 
However, the equivalence I'm struggling with is a hang-up for me. If the set B is a member of the power set of A wherein A is a member of family F, I don't see how it follows that there must be a value of the set A for which B is equivalent to the power set of A. There could be, but I see no logically sound reason to assume there is one.
 A: The set $\{P(A) \mid A \in \mathscr{F}\}$ is the set of powersets of $A$, where $A$ runs over all sets in the family $\mathscr{F}$. So if $B$ is an element of this set, it must be the powerset of some $A \in \mathscr{F}$, not just any old element in one of those power sets, that is, not just any generic subset of some $A \in \mathscr{F}$. 
I initially had the latter interpretation as well. There's an extra layer of "nesting" here that's tricky.
A: The thing is, writing
$$
\{\mathscr{P}(A)|A\in\mathscr{F}\}
$$
is a succinct way of writing
$$
\left\{E\in\mathscr{P}\left(\mathscr{P}\left(\bigcup\mathscr{F}\right)\right)\bigg|\exists A\in\mathscr{F},E=\mathscr{P}(A)\right\}.
$$
As you may well be aware of, when defining a set by describing the properties that its members have, you need to take those members in a set (here $\mathscr{P}(\mathscr{P}(\mathscr{\bigcup F}))$ was such a set). In general, one should always write something like $\{x\in X|\phi(x)\}$. This last set distinguishes the elements of $X$ that have the property $\phi(x)$. If you don't take the members from a set $X$, then you can get a contradiction, e.g. Russell's Paradox.
So,

If the set B is a member of the power set of A wherein A is a member of family F

would really be

If the set $B$ is a member of the family of power sets of sets in $\mathscr{F}$

So, the RHS of your equivalence just says that $B$ is one of those power sets.
A: We can start from your second example, rewriting it as :

$X=\{x \mid ∃n \in \mathbb N \ (x=n^2) \}$.

The formula inside the set-builder operator $\{ \ \mid \ \}$ can be thought as a sort of "procedure" : as long as $n$ "spans" the set $\mathbb N$, we get the output of the "equation" $x=n^2$ and "throw in" the result $x$ into the set $X$.
In the same way, for :

$\mathscr X = \{ B \mid ∃A \in \mathscr F \ (B = \mathscr P(A)) \}$

we have a family of sets : $\mathscr F$ (in place of $\mathbb N$). As long as $A$ spans it ($A$ is a set), we get the corresponding power-set $\mathscr P(A)$ (obviously, it exists, because $A$ is a set; and also $\mathscr P(A)$ is a set) and "throw it in" into the family $\mathscr X$ of power-sets defined by formula. 
