Suppose that $A$ is a set such that $|A| = m$. What is $| \{ X \subseteq \mathcal{P}(A) : |X| \le 1 \} |$?
I think that $| \{ X \subseteq \mathcal{P}(A) : |X| \le 1 \} | = | \mathcal{P}(A) | = 2^m$.
The set of all $X \subseteq \mathcal{P}$ is going to have more elements than $\mathcal{P}$. However, since we have the condition $|X| \le 1$, the set of all $X \subseteq \mathcal{P}$ will just be all of the elements in $\mathcal{P}$.
I would appreciate it if people could please review my solution and reasoning.
The question asks for the number of subsets of $\mathcal{P}(A)$ that are of cardinality $\le 1$. For any set $S$ the only subset of cardinality $0$ is the empty set, and there is one subset of cardinality $1$ for every $s\in S$ namely $\{s\}$. So the cardinality would be $|S|+1$.
Your answer is close to being correct. You just missed the empty set. The correct answer is $2^m+1.$
As for your last comment, yes $\emptyset\in\mathcal P(A)$. So both $\emptyset$ and $\{\emptyset\}$ are subsets of $\mathcal P(A).$