If $\kappa$ is an infinite cardinal and $\mathcal E$ is a maximal diverse subset of $\mathcal P(\kappa)$, then $|\mathcal E|=2^\kappa$.
Proof. Suppose $\mathcal E\subseteq\mathcal P(\kappa)$ is diverse and $|\mathcal E|\lt2^\kappa$; I will show that $\mathcal E$ is not maximal.
Let $\kappa=\bigcup_{\alpha\in\kappa}S_\alpha$ where the $S_\alpha$ are pairwise disjoint sets of cardinality $\kappa$. Define $f:\mathcal E\to\mathcal P(\kappa)$ by setting $f(X)=\{\alpha\in\kappa:|X\cap S_\alpha|=\kappa\}$.
Since $|\mathcal E|\lt2^\kappa$, $f$ is not surjective. Choose a set $A\subseteq\kappa$ which is not in the range of $f$, and let $Y=\bigcup_{\alpha\in A}S_\alpha$, so that $f(Y)=A$ and $Y\notin\mathcal E$.
If $X\in\mathcal E$ then $f(X)\ne f(Y)$, so there is some $\alpha\in\kappa$ such that either $|X\cap S_\alpha|\lt\kappa=|Y\cap S_\alpha|$ or else $|Y\cap S_\alpha|\lt\kappa=|X\cap S_\alpha|$, so $|(X\setminus Y)\cup(Y\setminus X)|=\kappa$, so $\mathcal E\cup\{Y\}$ is diverse, showing that $\mathcal E$ is not maximal.