meaning of $B \in \left \{ \mathcal P(A)|A \in \mathcal F \right \}$ 
$$B \in \left \{ \mathcal P(A)|A \in \mathcal F \right \}$$  

I am having a hard time understanding what this is saying. I know how to rewrite this to $\exists A \in \mathcal F(B=\mathcal P(A))$, which is the same meaning, but still I am confused about the meaning.
Translating the original: "the set $B$ is an element of the power set of $A$ such that $A$ is an element of the family of sets," correct?
I think that the family of sets is mainly confusing me. Also, If $B=\mathcal P(A)$, how can $B \in \mathcal P(A)$? I thought that because of Russell's paradox, a set cannot be an element of itself, i.e., $T \notin T$.
To sum up, I am super confused and super dense and appreciate help.  

$B$ is not an element of the powerset of $A,$ where $A\in \mathcal F.$ Rather, $B$ is a powerset of $A$ for some $A\in\mathcal F.$ As you wrote yourself, there exists $A\in\mathcal F$ such that $B=\mathcal P(A).$
When we say that $B\in\{\mathcal P(A)\mid A\in \mathcal F\},$ we're saying that $B$ is an element of the set $\{\mathcal P(A)\mid A\in\mathcal F\}.$ (As opposed to $B\in\mathcal P(A).$) Each element of this set is a powerset of $A$ for some $A\in\mathcal F.$ Hence $B$ is such a powerset.
For instance, if I told you, $x\in\{y\mid y\in\mathbb R \land y>0\},$ we'd know that $x=y>0$ and $x\in\mathbb R.$ We wouldn't try to conclude that $x\in y.$
Edit: You also seem to be thinking that there is some set in this description that necessarily contains itself, but this is not so.
