Is this Class\set theory equivalent to MK or NBG? Working in  bi-sorted \ FOL, where lower cases stand for sets and upper cases for classes; add all axioms of  ZFC-Ext.-Reg. written completely in lower case, and add axioms of Ext. and Reg. over all classes, as well as the following axioms:
Subsorting:$\forall x \ \exists Y: Y=x$
Elements:$\forall X \ \forall Y: X \in Y \to \exists z: z=X$
Class Abstraction schema: $\forall  \vec{Z} \ \exists X \ \forall y \ (y \in X \iff \phi(y, \vec{Z}))$; if $\phi(y, \vec{Z})$ is a formula not using $``X"$.
Now the question is about if this theory is equi-consistent to MK or is just a conservative extension over ZFC (i.e.; equi-consistent with NBG)?
 A: Edit: it turns out that this theory is just a conservative extension of ZFC.
Given a subtheory $J$ of ZFC - Ext - Reg, let $K_J$ be all axioms of $J$ in lowercase, plus extensionality, regularity, subsorting, elements, and class abstraction. Let $K = K_{ZFC - Ext - Reg}$ be the theory we're considering.
Working in the metatheory: Suppose $K \vdash \phi$ for some lowercase sentence $\phi$. Then take a finite subtheory $J$ of ZFC such that $K_J \vdash \phi$.
Working in ZFC: Take some ordinal $\alpha$ such that for all sentences $j \in J \cup \{\phi\}$, $j$ is absolute for $V_\alpha$. Write $M = V_\alpha$. This exists by the reflection theorem scheme. Then in particular, we see that $M$ is a transitive model of $J$.
$M$ will be the set of sets, and $P(M)$ the set of classes. Note that $P(M)$ is a transitive set.
All axioms of $J$, written in lower case, hold in this model, since they hold for $M$ by definition.
The axiom of subsorting holds.
The axiom of extensionality holds for $P(M)$ since $P(M)$ is transitive, and the axiom of regularity holds for the same reason.
The class abstraction scheme holds, since given some $\phi(y, \vec{Z})$ and some $\vec{Z}$, the set $\{z \in M \mid (M, P(M)) \models \phi(z, \vec{Z})\}$ exists.
This means that $(M, P(M)) \models K_J$. Therefore, $(M, P(M)) \models \phi$. Therefore, $M \models \phi$. But recall that $\phi$ is absolute for $M$. Therefore, $\phi$.
Back to the metatheory: So we have shown that if $K \vdash \phi$, then ZFC $\vdash \phi$. Thus, we see that $K$ is just a conservative extension of ZFC.
Thus, this theory is certainly not equivalent to MK unless ZFC is inconsistent, since MK proves that ZFC is consistent and thus proves that this new theory is consistent (since clearly we could take the metatheory for this whole exercise to be MK).
