Answer
a) $\mathcal{P}(\{ a \})$ $=\{\emptyset, \{a\}\}$.
b) $\mathcal{P}(\{a,b\})=\{\emptyset, \{a\}, \{b\}, \{a,b\}\}$.
c) $\mathcal{P}(\{\emptyset, \{\emptyset\}\})=\{\emptyset, \{\emptyset\}, \{\{\emptyset\}\}, \{\emptyset, \{\emptyset\}\}\}$.
Work Step by Step
Recall that for any set $A$, $\emptyset$ and the set $A$ itself are always subsets of $A$, so $\emptyset$ and $A$ are always in $\mathcal{P}(A)$.
A systematic way to list all the subsets of a set is to start with the empty set, then list the singletons, then list the two element sets, then the three element sets, and so on.
a) The subsets of $\{a\}$ are $\emptyset$, and $\{a\}$. So $\mathcal{P}(\{ a \})$ $=\{\emptyset, \{a\}\}$.
b) The subsets of $\{a,b\}$ are $\emptyset$, the singletons $\{a\}$ and $\{b\}$, and the two element set $\{a,b\}$. So $\mathcal{P}(\{a,b\})=\{\emptyset, \{a\}, \{b\}, \{a,b\}\}$.
c) Note that this is a two element set just like the set in part b), so this solution is analogous to the solution in part b). Thus the subsets of the two element set $\{\emptyset, \{\emptyset\}\}$ are $\emptyset$, the singletons $\{\emptyset\}$ and $\{\{\emptyset\}\}$, and the two element set $\{\emptyset, \{\emptyset\}\}$. So $\mathcal{P}(\{\emptyset, \{\emptyset\}\})=\{\emptyset, \{\emptyset\}, \{\{\emptyset\}\}, \{\emptyset, \{\emptyset\}\}\}$.
