Asaf states in his answer that if for all well-orderable sets $X$ the power set $\mathcal P(X)$ is also well-orderable, then the axiom of choice must hold. There is a proof of this left as exercise in Kunen's Set Theory, 2011 ed., exercise I.12.17, which comes with a hint, and is done as follows:
In $\mathsf{ZF}$, including the axiom of regularity of course, suppose that for any cardinal $\aleph$, $2^{\aleph}$ is well-orderable, then this implies that for any ordinal $\delta$, $\mathcal P(\delta)$ is well-orderable; as in $\mathsf{ZF}$ ,$\delta\thickapprox \aleph$, where $\aleph$ is the greatest cardinal with $\aleph\leq \delta.$
Let us prove by transfinite induction on $\gamma$ that $R(\gamma)$, the class of all sets of rank $\gamma$; which is a set, is well-orderable for all ordinals $\gamma$. As we are assuming $V=WF$ this will imply $\mathsf {AC}$.
Suppose $\gamma\neq 0 $ is such that the property holds for all $\alpha<\gamma$. If $\gamma=\alpha+1$ is successor, then as $R(\alpha)$ is well-orderable there is some ordinal $\delta$ such that $\delta\thickapprox R(\alpha)$, hence $R(\gamma)=\mathcal P(R(\gamma))\thickapprox \mathcal P(\delta)$, but as $\mathcal P(\delta)$ is well-orderable by hypothesis, $R(\gamma)$ is then well-orderable.
Now suppose $\gamma$ is limit. By Hartogs' Theorem; in $\mathsf{ZF}$, there is some cardinal $\kappa$ with $\kappa \npreceq R(\gamma)$. Fix a well-order $\sqsubset$ of $P(\kappa)$. Let us define by induction on $\alpha<\gamma$ an explicit well-order $\lhd_{\alpha}$ on $R(\alpha)$ for all $\alpha<\gamma$. Furthermore, let us construct each $\lhd_{\alpha}$ in such a manner that for any $\alpha<\beta<\gamma$ we have $\lhd_{\beta}\cap (R(\alpha)\times R(\alpha))\subseteq \lhd_{\alpha}$.
Let $\lhd_0$ be the canonical well-order of $R(0)$. Suppose we have defined $\lhd_{\alpha}$ for all $\alpha<\beta$ for some $\beta\leq \gamma$, meeting the conditions above. There are two cases:
Case $\beta=\alpha+1$. Since $\kappa\npreceq R(\alpha)$, we must have that $\bf{type}$$(R(\alpha),\lhd_{\alpha})<\kappa.$ Let $f_{\alpha}:(R(\alpha),\lhd_{\alpha})\rightarrow \kappa$ be the canonical embedding, the one that maps $(R(\alpha),\lhd_{\alpha})$ into an initial segment of $\kappa$. Then $f_{\alpha}[R(\alpha)]=\mu$ for some $\mu<\kappa$, so that $f^{-1}_{\alpha}[ \ ]:P(\mu)\rightarrow P(R(\alpha))$ is a bijection, but $P(\mu)\subseteq P(\kappa)$. Hence there is an explicit well-ordering of $P(R(\alpha))$ via $f^{-1}_{\alpha}[ \ ]$ and the well-order $\sqsubset$ on $P(\kappa)$. Let $\lhd_{\beta}$ be this well-order. Then clearly for any $\mu<\beta$ we have $\lhd_{\beta}\cap (R(\mu)\times R(\mu))=\emptyset$.
Suppose $\beta$ is limit. Since for any $\alpha<\mu<\beta$ we have $\lhd_{\mu}\cap R(\alpha)^2\subseteq \lhd_{\alpha}$, it follows that if we put $\lhd_{\beta}=\bigcup_{\alpha<\beta}\lhd_{\alpha}$, $\lhd_{\beta}$ is a well-order on $R(\beta)=\bigcup_{\alpha<\beta}R(\alpha)$, and also for any $\mu<\beta$, $\lhd_{\beta}\cap R(\mu)^2\subseteq \lhd_{\mu}.$
Hence by transfinite induction we obtained an explicit well-order $\lhd_{\gamma}$ of $R(\gamma)$. Hence $R(\gamma)$ is well-orderable for all ordinals $\gamma$, and therefore $\mathsf{AC}$ holds.
