Lets $\mathsf{MK^-}$ Be the theory "$\mathsf{MK}-\text{Limitation of size}+\text{Subsets}-\text{Union}$", removal of axiom of limitation of size and putting the axiom of subsets (the axiom asserting that every subclass of a set is a set) instead of it would make it possible for some classes to be equinumerous to sets and yet not being sets, since it is known that the axiom of subsets does not imply the assertion that every class that is equinumerous to a set is a set. So is it possible to have a model $M$ of $\mathsf{MK^-}$ such that we have all of the followings:

$M \vdash \exists P (\text {$P$ is a proper class} \wedge P < V)$
$M \vdash \forall P (\text{$P$ is a proper class} \wedge P < V \implies \exists x (x\in V \wedge \text{$P$ is equinumerous to x}))$

where "<" denotes "strict subnumerousity" defined in the customary manner; and $V$ is the class of all sets. Given that it is clear that $M$ cannot satisfy global choice nor can it satisfy closure of replacement over sets. This question is about size of proper classes and to what extent that can be shared with size of sets in absence of Replacement and global choice.

Also related to this is the following question:

If the above is possible then can we add the following:

$M \vdash \forall P (\text{$P$ is a proper class} \wedge P < V \implies \exists x (x=\{\{y\}| y \in P\} \wedge x \in V))$

The special context of those questions raised when I was investigating an alternative to the axiom of limitation of size of $\mathsf{MK}$. A version that I've lately posted to FOM thread is referred to by the following link:

http://www.cs.nyu.edu/pipermail/fom/2016-September/020073.html

This version proves all axioms of Empty set, Pairing, Power, Separation (subsets), Replacement, Global choice and Set union, so it is more powerful than the usual limitation of size axiom.

I've been trying to further simplify this axiom to the following:

$\forall x (x \in V \iff U(x) < V)$

However I couldn't prove Replacement nor union, so I was left with the above situation (i.e. theory $\mathsf{MK^-}$); so if the possibilities that I've asked about specifically above were inconsistent then this mean that the axiom can be simplified to the above version.