Can a Vopenka cardinal be supercompact?


I asked a [weaker question](https://mathoverflow.net/questions/400732/is-vopenkas-principle-ord-has-the-tree-property-consistent) on here before. Unfortunately, I don't know enough set theory to see whether the positive answer there generalizes to a positive answer here.



Can a Vopenka cardinal be supercompact?
If $\kappa$ is almost huge with target $\lambda,$ then $V\_{\lambda}$ thinks that $\kappa$ is a supercompact Vopenka cardinal.


I'll take for granted the standard facts about almost huge cardinals listed here: <https://neugierde.github.io/cantors-attic/Huge>


In particular, $\kappa$ is Vopenka (in $V$), and this is just an assertion about $V\_{\kappa+1}$ so $V\_{\lambda}$ also sees that $\kappa$ is Vopenka. Also $\kappa$ is $\theta$-supercompact for every $\theta<\lambda,$ i.e. every $\mathcal{P}\_{\kappa}(\theta)$ has a normal fine measure. Again these facts are all absolute down to $V\_{\lambda},$ so $V\_{\lambda}$ thinks that $\kappa$ is supercompact.


Also by a standard elementary embedding argument, these facts about $\kappa$ and $\lambda$ imply that there are stationarily many $\alpha<\kappa$ such that $V\_{\kappa}$ believes $\alpha$ to be a supercompact Vopenka cardinal.


