An equality of sets involving two directions LEt $A,B$ be sets. Prove that $A \subseteq B \iff P(A) \subseteq P(B) $.

Attempt:

First, take any element of $P(A)$, say $Y$, we know by definition that $Y \subseteq A$ and so $Y \subseteq B$ (given). Therefore, by definition $Y \in P(B)$. and we have the result.

Conversely, suppose $x \in A$, then $\{x\} \in P(A)$ and so $\{ x \} \in P(B)$. (Can we conclude that $x \in B$) ?
Yes, that looks right.

For the single remaining step, once you know that $\{x\}\in\mathcal P(B)$, this is the same as $\{x\}\subseteq B$, which again means that $x\in B$, as needed.
I am afraid you seem to be getting a little confused on notation $\in$ is used to denote an element belonging to a set and $\subset$ is when saying a set is a subset of another set.

Anyway to show the converse assume the right hand side of the implication and let $x\in A$ then $x \in P(A)$ indeed taking the subset with just the element $x$ then $x \in P(B)$ and since $x$ is just a single element if $x \in P(B)$ then it follow $x \in B$ and we have proved the converse.

The first direction is fine one you sort out the notation. I think just need to make sure you are clear on notation.