Recall the definition of cardinal definable sets, to re-iterate:

$Define: X \text { is cardinal definable} \iff \\\exists \text { cardinal } \kappa \, \exists \text { cardinals } \lambda_1,.., \lambda_n <^\rho \kappa \ \exists \phi : \\ X=\{ y \in V_{\rho(\kappa)} \mid \phi^{V_{\rho (\kappa)}} (y,\lambda_1,..,\lambda_n)\}$

Where: $\lambda_i <^\rho \kappa \iff \rho(\lambda_i) < \rho(\kappa)$, and $\rho$ is the rank function; and "cardinal" is defined after Scott's as an equivalence class under bijection of sets of the lowest possible rank.

That every set is cardinal definable is proved consistent with failure of choice (see here), on the other hand it is also proved consistent with choice (see here).

Now, working in $\sf ZF$, if we say that every cardinal definable set admits a choice function, would that entail full $\sf AC$?

If we work in $\sf ZF-Reg.$, would cardinal definable choice imply $\sf AC$?