If A is a set and P(A) its power set, then P(A) contains all the elements in A. P(A) also contains the subsets of A, which A doesn't necessarily contain.
So, does P(A) contain a copy of all the elements in A, and not the same elements?
Hence, if A = {x} is a singleton, B = {x} ⊆ P(A), then A and B are not the same element, despite A = B. Can this be clarified to me?
I'm assuming here that "$P(A)$" means "the powerset of $A$."
Your claim that in general $A\subseteq P(A)$ is false. For simplicity, let's work in naive set theory (although everything I say is also true in, say, ZFC). Then:
The powerset of the set $A=\{$apple$\}$ is $P(A)=\{\emptyset, \{$apple$\}\}$.
However, apple isn't an element of $P(A)$: $\emptyset\not=$ apple and $\{$apple$\}\not=$ apple.
I think the key point is this last bit: in general, we won't have $x=\{x\}$. (Indeed, in ZFC we never will have this, by the axiom of regularity!) The ur-example is $\{\emptyset\}$ versus $\emptyset$ (one of these isn't empty!).
What is true is that there is a natural injection of $A$ into $P(A)$: for $a\in A$, the set $\{a\}$ is an element of $P(A)$, and the map $A\rightarrow P(A):a\mapsto \{a\}$ (the "singleton map") is injective. That is, for each element $a$ of $A$, the powerset $P(A)$ contains something corresponding to the object $a$, namely $\{a\}$. But this is not the same thing as the object $a$.