Let's assume the actual question is to find the power set of $S=\{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}$. Using Von Neumann's definition of the natural numbers, this is equivalent to finding the power set of $S=\{0,1,2\}$, where $0=\emptyset$, $1=\{0\}$, and $2=\{0,1\}$. The power set is then
$$\{\emptyset,\,\{0\},\,\{1\},\,\{2\},\,\{0,1\},\,\{0,2\},\,\{1,2\},\,\{0,1,2\}\}$$
Just change the $2$'s to $\{0,1\}$, then the $1$'s to $\{0\}$, then the $0$'s to $\emptyset$--and you are done!
You will end up with a royal mess, of course. I found it difficult enough to format the easier form of the power set above.
Pick out the number $2$ so you are left with the
set $\{1,3,4,...,n\}$. Now you construct all subsets of
this set and add a $2$ to each of these subsets.
As $\{1,3,4,...,n\}$ is a set of $n-1$ elements
it has in total $2^{n-1}$ subsets and by your
construction you have in total $2^{n-1}$ subsets
of $\{1,2,3,...,n\}$ that contains a $2$.
Best Answer
Suppose $A=\{a_1,a_2,\dots,a_n\}$, then $$\mathcal P(A)=\{\emptyset,\overbrace{\{a_1\},\{a_2\},\dots,\{a_n\}}^{\binom{n}{1} \text{ sets}},\overbrace{\{a_1,a_2\},...}^{\binom{n}{2}\text{ sets}},\overbrace{\{a_1,a_2,a_3\}}^{\binom{n}{3} \text{ sets}},...,A\}.$$
Recall that $\binom{n}{k}$ tells you the number of sets you can make with $k$ elements from a set of $n$ elements.
Then if we add everything:
$$|\mathcal P(A)|=\sum\limits_{i=0}^n\binom{n}{i}=\sum\limits_{i=0}^n\binom{n}{i}(1^i)(1^{n-i})\stackrel{\text{by Binomial Theorem}}{=}(1+1)^n=2^n.$$ Why do we sum $\binom{n}{0}$ and $\binom{n}{n}$? Because of the number of sets with no elements (this is the empty set) and the number of sets with $n$ elements (this is $A$). Therefore you are interested in:
$$\sum\limits_{i=0}^2\binom{2^n}{i}.$$