This is an old exam question from "Diskrete Mathematik" at ETH Zurich
$\mathcal{P}(A)$ denotes the Powerset, which is the set of all subsets of A:
$\mathcal{P}(A) := \{S|S\subseteq A \}$.
I believe that $\{\emptyset, \mathcal{P}(\emptyset), \{\emptyset\} \} $ since $\mathcal{P}(\emptyset) = \{\emptyset\} $ can be reduced to $\{\emptyset, \{\emptyset\} \} $ and thus the cardinality of said set would be 2. However, the inofficial solution is 3.
The power set of a finite set with cardinality $x$ has cardinality $2^x$. As you pointed out, $\{\emptyset, \mathcal{P}(\emptyset), \{\emptyset\} \}$ can be simplified down to $\{\emptyset, \{\emptyset\} \}$, which is equivalent to $\mathcal{P}(\{\emptyset\})$. Using the property I described initially, we end up with,
$$ 2^{|\{\emptyset\}|} = 2^1 = 2. $$
Therefore, the correct solution is $2$ and not $3$ as the unofficial solution suggested.