I will give a partial answer for models of $NBG$ in which local choice ($AC$) holds, namely:
Theorem. There is a model $M$ of $ZFC$ whose natural expansion $(M,\cal{A})$ to a model of $NBG$ (where $\cal{A}$ is the collection of all parametrically definable classes of $M$) has at least 3 bijection-inequivalent classes, namely:
(1) the class $Ord$ of ordinals,
(2) the class $\cal{P}$$(Ord)$ of all subsets of ordinals, and
(3) the class $V$ of sets.
Moreover, it is a theorem of $NBG$ plus $AC$ that there is a bijection between $V$ and $\cal{P}(\cal{P}$$(Ord))$ (the class of all subsets of $\cal{P}$$(Ord))$.
Proof. Let $M$ be a model of $ZFC$ which has a definable class of pairs with no no choice function (see, e.g., the answers given by Hamkins and myself to this MO question).
It is clear that there is no injection from $V$ into $Ord$ that is canonically coded in $\cal{A}$.
We next observe that there is no injection from $V$ into $\cal{P}$$(Ord)$ that is coded in $\cal{A}$. This is because there is an obvious definable linear ordering of $\cal{P}$$ (Ord)$ (by viewing each subset of ordinals as a binary sequence), and the existence of such an injection would contradict the fact that there is no definable choice function on a class of pairs.
Finally, to show that there is no injection from $\cal{P}$$(Ord)$ into $Ord$ in our model of $NBG$, we take advantage of the axiom of choice within $M$. More specifically, recall that in the presence of the axiom of choice, every set is definable from a subset of ordinals, the idea being: given any set $x$, let TC($x$) be the transitive closure of {$x$}. Then by AC there is bijection $f$ between TC($x$) and some ordinal $\alpha$. This allows us to copy the $\in$ relation on TC($x$) to a binary relation $r$ on $\alpha$, and then we can use a canonical pairing function on ordinals to code $r$ as some subset $s$ of ordinals. Then $x$ is definable from $s$ via "$x$ is the top element of the transitive collapse of $r$".
Note that the above procedure gives rise to a definable surjection $F$ from $\cal{P}$$(Ord)$ onto $V$: given a subset $s$ of ordinals, first decode it as a binary relation $r$ on ordinals, and then ask whether $r$ is an extensional, well-founded relation with a top element. If no, let $F(s)$ be $0$, and if yes, then let $F(s)$ be the top element of the transitive collapse of $r$.
Now if there exists a definable injection $F$ from $\cal{P}$$(Ord)$ into $Ord$, then in light of the fact that $\cal{P}$$(Ord)$ would then be definably well-ordered, the above map $F$ could be "inverted" by an injection $G$ from $V$ into $\cal{P}$$(Ord)$, where $G(x)$ is the first $s$ such that $F(s)=x$. This, in turn would enable us to definably well-order $V$, contradiction.
Finally, to verify the last claim of the theorem: the coding explained above allows us to define an injection $H$ from $V$ into $\cal{P}(\cal{P}$$(Ord))$, namely $H(x)$ is the collection of all subsets $s$ of $\kappa$, where $\kappa$ is the cardinality of the transitive closure of {$x$}, such that when $s$ is canonically decoded as a relation $r$ on $\kappa$, then $r$ is an extensional well-founded relation with a top element such that $x$ is equal to the top element of the transitive closure of $r$.
Therefore, since the class version of the Schröder-Bernstein theorem holds in models of $NBG$, there is a bijection between $V$ and $\cal{P}(\cal{P}$$(Ord))$ in models of $NBG$ in which the (local) axiom of choice holds.
