I don't know the truth value of this statement so this is just my own attempt. I am attempting to prove this in NBG set theory.
Let $X \in \textit{P(On)}$ By the definition provided in my notes $X \in \textit{On}$ iff $X$ is well ordered by the $\in$-relation and is what my notes call 'full'.
Fullness is the property whenever $y \in X$ $\implies$ $y \subseteq X$
Assume $0\neq X$ since if $0 = X\in \textit{P(On)}$ $\implies X\in \textit{On}$ follows trivially.
So $0\neq X \in \textit{P(On)}$
First to show fullness:
It is taken as a proven theorem that $X$ is full iff $\ \bigcup{X} \subseteq X$
Let $y \in \bigcup{X}$ $\implies y\in z$ for some $z \in X$
Since $X \in \textit{P(On)}$ $\implies$ $X\subseteq \textit{On}$
So $z \in X \subseteq \textit{On} \implies z\in \textit{On}$
It is an established fact in my notes that $\textit{On}$ is full $\implies$
$y \in z \subseteq \textit{On}$ $\implies y\in \textit{On}$
So $y \subseteq z \in X \implies y \in X$ (!!!)
I'll stop here since I am uncertain about the validity of the final step.
This would allegedly prove the fullness of $X$. Well ordering would follow easily from this.
You're right to be uncertain: that last step doesn't work. There's no reason that $y\subseteq z\in X$ should imply $y\in X$.
In fact, it is not true that every element of $P(On)$ is full. For instance, $\{\emptyset\}\in On$ so $\{\{\emptyset\}\}\in P(On)$, but $\{\{\emptyset\}\}$ is not full.
(Also, by the way, the standard term for "full" is "transitive".)