Answer
a) No
b) Yes
c) No
d) Yes
Work Step by Step
a) No. The power set of any set must contain the empty set. Thus for any set $A$, $\mathcal{P}(A) \neq \emptyset$.
b) Yes. This is the power set of $\{a\}$. Note the subsets of $\{a\}$ are $\emptyset $ and $\{a\}$, so $\mathcal{P}(\{a\}) = \{\emptyset , \{a\}\}$.
c) No. To see this, suppose $\mathcal{P}(A) = \{\emptyset , \{a\}, \{ \emptyset , a\}\}$ for some set $A$, then $\{\emptyset , a\}$ would be a subset of $A$. This means every element in $\{\emptyset , a\}$ is in $A$. Hence $\emptyset \in A$, which means $\{\emptyset \} \subseteq A$, and, therefore, $\{ \emptyset \} \in \mathcal{P}(A)$. But since $\mathcal{P}(A) = \{ \emptyset , \{a\}, \{ \emptyset , a\} \}$, we can see $\{ \emptyset \} \notin \mathcal{P}(A)$. Thus $\{ \emptyset , \{a\}, \{ \emptyset , a\} \} \neq \mathcal{P}(A)$ for any set $A$.
d) Yes. This is the power set of $\{a,b\}$. To see this, notice the subsets of $\{a,b\}$ are $\emptyset$, $\{a\}$, $\{b\}$, and $\{a,b\}$. Thus $\mathcal{P}(\{a,b\}) = \{\emptyset , \{a\}, \{b\}, \{a,b\}\}$.
