If $E$ is finite then $\mathcal P_\kappa(E)$ is finite, and has strictly more sets than $[E]^\kappa$, so the cardinality must be different.
If $E$ is infinite then $|\mathcal P_\kappa(E)|\leq|E^\kappa|=|E|^\kappa$, because the function sending $f\in E^\kappa$ to its range is surjective (sans the empty set, but one element is not important in the infinite case anyway). As Proving that for infinite $\kappa$, $|[\kappa]^\lambda|=\kappa^\lambda$ show, $[E]^\kappa$ has cardinality $|E|^\kappa$. Therefore equality ensues.
Note that it is obvious that $\kappa\leq|E|$ otherwise $\mathcal P_\kappa(E)$ is $\mathcal P(E)$, whereas $[E]^\kappa$ is empty.