Is there such an expression as $\displaystyle \forall B \mathcal,{P}(B)$?
It's very hard for me imagine there would be more than one $\displaystyle B$ in $\displaystyle \mathcal{P}(B)$.
That's a formula, though it's an odd one. It seems not what you intend.
What you probably mean is
$\displaystyle \forall A \in \mathcal{P}(B)(A \subseteq B)$
No.
$\displaystyle \{Y:Y\subseteq B\}$
is a term; it's a "name"; it names a particular set; it doesn't make an assertion.
$\displaystyle \forall Y \in \mathcal{P}(B)(Y \subseteq B)$
is not a term, but rather it's a formula; it makes an assertion (once we agree on what 'B' stands for, or, alternatively, once we decide it should be taken in the sense of a universal generalization on 'B').
You very much need to get that logic book I recommended, as soon as possible; it will get you on the right track to writing correct symbolizations.
I am supposed to provide an example of a set $\displaystyle A$ that satisfies$\displaystyle \forall B P(B)(A\subseteq B)$.
$\displaystyle A= \varnothing$ seems to fit the expression, but I know there is only one $\displaystyle B$ in $\displaystyle P(B)$
If B={1,2}, P(B)={$\displaystyle \varnothing$ {1},{2},B}. I wondering why $\displaystyle \forall B$ is used. It doesn't make sense.
I don't recognize that as a formula. Yes, it doesn't make sense. I can't find an A that satisfies it, because it's not even a formula.
The earlier formula you gave was
$\displaystyle \forall B \in \mathcal{P}(B)(A \subseteq B)$
That is equivalent to:
For all B (If B is a subset of B, then A is subset of B).
An A that satisfies that is B itself, or any other A that is a subset of B, depending on what B is.
Though maybe you've not rendered the formula correctly (it is rather odd)?
It surely is a typo of some sort.
I know of several books that use this notation, $\displaystyle P(B:A)$.
It stands for $\displaystyle A\in P(B)$ and $\displaystyle P(B:A)=\{X\in P(B):A\subseteq X\} $.
All subsets of B that have A as a subset.
Is that close to what you think it may mean?