Let $X\neq\emptyset$ be a set and let $\mu:{\cal P}(X)\to [0,1]$ be a probability measure. Is there a probability measure $$\bar{\mu}:{\cal P}({\cal P}(X))\to [0,1]$$ with the following property?
For all $S\subseteq X$ we have $\bar{\mu}({\cal P}(S)) =\mu(S)$.
For notational and conceptual simplicity, assume that $X$ and $P(X)$ are disjoint.
As Bugs Bunny implicitly suggested: Let $\mathcal E = \{\{n\}: n \in X \}$ be the set of singletons. This is a subset of $P(X)$. The measure $\bar\mu$ will concentrate on this set.
For any $\mathcal A \subseteq P(X)$, we will have $\bar \mu(\mathcal A)=\bar \mu(\mathcal A\cap\mathcal E)$; note that $\mathcal A\cap\mathcal E$ is morally the same as a subset of $X$, so $\mu$ will measure it.
Formally, let $S_{\mathcal A} = \{n\in X: \{n\}\in \mathcal A\}$ be the ``singleton support'' of $\mathcal A$. Define $\bar \mu (\mathcal A) := \mu(S_{\mathcal A})$.
If $|X|>2$, then there is a single solution: $\overline{\mu}(\{x\})=\mu(x)$ and $\overline{\mu}(S)=0$ for any other set $S$.
Indeed, let $a=\overline{\mu}(\emptyset)$. By the condition for $S=\{x\}$, $\overline{\mu}(\{x\}) =\mu (x)-a$. By the condition for $S=\{x,y\}$ with $x\neq y$, $\overline{\mu}(\{x,y\}) =a$.
Now if $|X|>2$, there are at least $1+|X|$ two-element subsets. Thus, $\sum_{|S|\leq 2} \overline{\mu}(S) \geq a+ \sum_{x\in X}\mu (x) =1+a$. This forces $a=0$.
Now it is easy to show that for all larger subsets $\overline{\mu}(S)=0$.
