Over ZF yes, it does.

Let $X$ be any family of nonempty sets, and let $κ$ be cardinal such that $ρ(κ)>ρ(X)$, then $V_{ρ(κ)}\setminus\{∅\}$ is cardinal definable, hence has a choice function that induce a choice function on $X$.

In ZF-Regularity Scott trick doesn't work anymore, in fact Lévy has shown that it is consistent with ZF-Regularity to assign to each set a cardinality in any way (that is: it is consistent that there is no $C:V→V$ such that $C(x)=C(y)⇔|x|=|y|$ (even with parameters)).

In e.g. ZF-Regularity+AFA were you do have possible definition of cardinality for every set, similar idea to what I wrote above works: You start with with a set $X$, and define $X'=X\cap WF$ and $X^∘=X\setminus X'$, by the above $X'$ has a choice function, for $X^∘$ define $Y$ to be the class of accessible pointed graphs who are pictures for elements is in $Y$. Let $f:Y∩WF→X^∘$ be a function that sends $y$ to the unique element in $X^∘$ that $y$ is it's picture.

Note that $f''(Y∩WF)=X^∘$, and in fact that there exists a set $Z\subseteq Y∩WF$ such that $f''Z=X^∘$ (We don't need choice to get this $Z$, we can use Scott's trick, because $Y∩WF\subseteq WF$).

From the above $Z$ and $\bigcup Z$ are well-orderable, so let $x\in X^∘$, and $(G,E)∈Z$ be the minimal graph with the unique decoration $π$ such that $x\in\pi'' G$, and let $p$ be the minimal element (in $\bigcup Z$) such that such that $p E\pi^{-1}(x)$, and so $π(p)∈x$ and we have a choice function on $X=X'⊔X^∘$

Over ZF yes, it does.

Let $X$ be any family of nonempty sets, and let $κ$ be cardinal such that $ρ(κ)>ρ(X)$, then $V_{ρ(κ)}\setminus\{∅\}$ is cardinal definable, hence has a choice function that induce a choice function on $X$.

In ZF-Regularity Scott trick doesn't work anymore, in fact Lévy has shown that it is consistent with ZF-Regularity that there is no assignment to each set a cardinality in any way (that is: it is consistent that there is no $C:V→V$ such that $C(x)=C(y)⇔|x|=|y|$ (even with parameters)).

In e.g. ZF-Regularity+AFA were you do have possible definition of cardinality for every set, similar idea to what I wrote above works: You start with with a set $X$, and define $X'=X\cap WF$ and $X^∘=X\setminus X'$, by the above $X'$ has a choice function, for $X^∘$ define $Y$ to be the class of accessible pointed graphs who are pictures for elements is in $Y$. Let $f:Y∩WF→X^∘$ be a function that sends $y$ to the unique element in $X^∘$ that $y$ is it's picture.

Note that $f''(Y∩WF)=X^∘$, and in fact that there exists a set $Z\subseteq Y∩WF$ such that $f''Z=X^∘$ (We don't need choice to get this $Z$, we can use Scott's trick, because $Y∩WF\subseteq WF$).

From the above $Z$ and $\bigcup Z$ are well-orderable, so let $x\in X^∘$, and $(G,E)∈Z$ be the minimal graph with the unique decoration $π$ such that $x\in\pi'' G$, and let $p$ be the minimal element (in $\bigcup Z$) such that such that $p E\pi^{-1}(x)$, and so $π(p)∈x$ and we have a choice function on $X=X'⊔X^∘$

Over ZF yes, it does.

Let $X$ be any family of nonempty sets, and let $κ$ be cardinal such that $ρ(κ)>ρ(X)$, then $V_{ρ(κ)}\setminus\{∅\}$ is cardinal definable, hence has a choice function that induce a choice function on $X$.

In ZF-Regularity Scott trick doesn't work anymore, in fact Lévy has shown that it is consistent with ZF-Regularity to assign to each set a cardinality in any way (that is: it is consistent that there is no $C:V→V$ such that $C(x)=C(y)⇔|x|=|y|$ (even with parameters)).

In e.g. ZF-Regularity+AFA were you do have possible definition of cardinality for every set, similar idea to what I wrote above works: You start with with a set $X$, and define $X'=X\cap WF$ and $X^∘=X\setminus X'$, by the above $X'$ has a choice function, for $X^∘$ define $Y$ to be the class of accessible pointed graphs who are pictures for elements is in $Y$. Let $f:Y∩WF→X^∘$ be a function that sends $y$ to the unique element in $X^∘$ that $y$ is it's picture.

Note that $f''(Y∩WF)=X^∘$, and in fact that there exists a set $Z\subseteq Y∩WF$ such that $f''Z=X^∘$.

From the above $Z$ and $\bigcup Z$ are well-orderable, so let $x\in X^∘$, and $(G,E)∈Z$ be the minimal graph with the unique decoration $π$ such that $x\in\pi'' G$, and let $p$ be the minimal element (in $\bigcup Z$) such that such that $p E\pi^{-1}(x)$, and so $π(p)∈x$ and we have a choice function on $X=X'⊔X^∘$

Over ZF yes, it does.

Let $X$ be any family of nonempty sets, and let $κ$ be cardinal such that $ρ(κ)>ρ(X)$, then $V_{ρ(κ)}\setminus\{∅\}$ is cardinal definable, hence has a choice function that induce a choice function on $X$.

In ZF-Regularity Scott trick doesn't work anymore, in fact Lévy has shown that it is consistent with ZF-Regularity to assign to each set a cardinality in any way (that is: it is consistent that there is no $C:V→V$ such that $C(x)=C(y)⇔|x|=|y|$ (even with parameters)).

In e.g. ZF-Regularity+AFA were you do have possible definition of cardinality for every set, similar idea to what I wrote above works: You start with with a set $X$, and define $X'=X\cap WF$ and $X^∘=X\setminus X'$, by the above $X'$ has a choice function, for $X^∘$ define $Y$ to be the class of accessible pointed graphs who are pictures for elements is in $Y$. Let $f:Y∩WF→X^∘$ be a function that sends $y$ to the unique element in $X^∘$ that $y$ is it's picture.

Note that $f''(Y∩WF)=X^∘$, and in fact that there exists a set $Z\subseteq Y∩WF$ such that $f''Z=X^∘$ (We don't need choice to get this $Z$, we can use Scott's trick, because $Y∩WF\subseteq WF$).

From the above $Z$ and $\bigcup Z$ are well-orderable, so let $x\in X^∘$, and $(G,E)∈Z$ be the minimal graph with the unique decoration $π$ such that $x\in\pi'' G$, and let $p$ be the minimal element (in $\bigcup Z$) such that such that $p E\pi^{-1}(x)$, and so $π(p)∈x$ and we have a choice function on $X=X'⊔X^∘$