Assuming $P(A)$ is the power set of $A$, would this be a correct counterexample for the statement that $P(A-B) = P(A) - P(B)$?
Let $A = \{2, 3\}$ and $B = \{2\}$, therefore $C = \{3\}$.
$P(A-B) = P(C) = \{\{3\}, \{\varnothing\}\}$
\begin{align*} P(A) - P(B) &= \{\{2, 3\}, \{2\}, \{3\}, \{\varnothing\}\} - \{\{2\}, \{\varnothing\}\} \\ &= \{\{2, 3\}, \{3\}\} \end{align*}
As shown, $P(A-B)$ contains the empty set while $P(A) - P(B)$ does not and as such:
$$P(A-B) \ne P(A) - P(B)$$
Comments from Cameron Williams and Asaf Karagila were incorporated in this answer.
Yes that would work.
In fact, $P(A) - P(B)$ never contains the empty set (as $\varnothing \in P(B)$), so any $A$ and $B$ suffice (even $A = B = \varnothing$).