Multiple choice produces maximal antichains

If $P=\mathrm{\varnothing }$, its empty subset is maximal. Otherwise use MC to fix $s(U)$ finite nonempty in every nonempty $U\subseteq P$. The minimal elements $m(U)$ of $s(U)$ form a nonempty finite antichain: descent in a finite strict poset terminates, and two minimal points cannot be comparable.

Recurse for $\xi <h(\mathcal{P}(P))$. Given earlier antichains, let $U_{\xi }$ be the points outside their union incomparable with every point in that union. Set $A_{\xi }=m(U_{\xi })$ if it is nonempty, and $A_{\xi }=\mathrm{\varnothing }$ otherwise. Earlier and later nonempty stages are disjoint and mutually incomparable.

If every stage were nonempty, $\xi \mapsto A_{\xi }$ would inject $h(\mathcal{P}(P))$ into $\mathcal{P}(P)$ because the stages are disjoint. Thus some $U_{\xi }$ is empty. The union of the stages before the first such index is an antichain to which no point can be added.