Cardinality of the power set

1. The subsets of $E_{p+1}$ split into: (a) those contained in $E_p$, of which there are $|\mathcal{P}(E_p)|$; (b) those of the form $S\cup\{a_{p+1}\}$ with $S\subseteq E_p$, also $|\mathcal{P}(E_p)|$ in number. Hence $|\mathcal{P}(E_{p+1})|=2|\mathcal{P}(E_p)|$.

2. Multiplying $|\mathcal{P}(E_{p+1})|=2|\mathcal{P}(E_p)|$ for $p=0,\ldots,n-1$:
$$|\mathcal{P}(E_n)|=2^n|\mathcal{P}(E_0)|.$$
Since $E_0=\emptyset$, $\mathcal{P}(E_0)=\{\emptyset\}$, so $|\mathcal{P}(E_0)|=1$, and $|\mathcal{P}(E_n)|=2^n$.

Conclusion. $|\mathcal{P}(E_n)|=2^n$.