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$\def\rk{\mathop{\rm rank}}\def\ZZ{{\mathbb Z}}$ We need to estimate the number of expansions of the form $A:=A_1-A_2=C_1-C_2$ with $\rk C_i\leq n-k$ (then $B=A_1-C_1=A_2-C_2$). Since the multiplication by a non-degenerate matrix does not change anything, this depends only on $\rk A$.

Assume, for instance, that $A$ is non-degenerate. Fix two $(n-k)$-dimensional subspaces $V_1,V_2$ of the space of rows $V=\ZZ_p^n$ with $\dim V_i=n-k$; counting the bases, we obtain that there are there are $$N=\displaystyle \frac{(p^n-1)(p^n-p)\dots (p^n-p^{n-1})} {(p^{n-2k}-1)(p^{n-2k}-p)\dots(p^{n-2k}-p^{n-2k-1})\cdot \bigl((p^{n-k}-p^{n-2k})(p^{n-k}-p^{n-2k+1})\dots(p^{n-k}-p^{n-k-1})\bigr)^2}$$ such pairs of subspaces.

Denote by $C_i^j$ the $j$th row of $C_i$. Now let us count all the pairs $(C_1,C_2)$ such that $\mathop{\rm span} (C_i^1,\dots,C_i^n)\subset C_i^1,\dots,C_i^n)\subseteq V_i$. Let $V'=V_1\cap V_2$. Then $C_i^j\mod V'$ is determined uniquely, hence we have $p^{n-2k}$ variants for each $C_1^j$, and $C_2$ is reconstructed from $C_1$. Thus we have $p^{n(n-2k)}$ pairs.

In total, we get $N\cdot p^{n(n-2k)}$ pairs. In fact, this is an upper bound, since the rank of $C_i$ may be less than $p-k$, and in this case one pair will correspond to several pairs $(V_1,V_2)$.

You may easily obtain a bound for $N$, but it would be better to know the relation between $p$ and $n$...

If $A$ is degenerate, then this bound should increase, since we just need $V_1+V_2$ to contain the space generated by the rows of $A$. On the other hand, we may restrict ourselves to the case when $(V_1+V_2)\mod V'=\mathop{\rm span}(A^1,\dots,A^n)\mod V'$.

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$\def\rk{\mathop{\rm rank}}\def\ZZ{{\mathbb Z}}$ We need to estimate the number of expansions of the form $A:=A_1-A_2=C_1-C_2$ with $\rk C_i\leq n-k$ (then $B=A_1-C_1=A_2-C_2$). Since the multiplication by a non-degenerate matrix does not change anything, this depends only on $\rk A$.

Assume, for instance, that $A$ is non-degenerate. Fix two $(n-k)$-dimensional subspaces $V_1,V_2$ of the space of rows $V=\ZZ_p^n$ with $\dim V_i=n-k$; counting the bases, we obtain that there are there are $$N=\displaystyle \frac{(p^n-1)(p^n-p)\dots (p^n-p^{n-1})} {(p^{n-2k}-1)(p^{n-2k}-p)\dots(p^{n-2k}-p^{n-2k-1})\cdot \bigl((p^{n-k}-p^{n-2k})(p^{n-k}-p^{n-2k+1})\dots(p^{n-k}-p^{n-k-1})\bigr)^2}$$ such pairs of subspaces.

Denote by $C_i^j$ the $j$th row of $C_i$. Now let us count all the pairs $(C_1,C_2)$ such that $\mathop{\rm span} (C_i^1,\dots,C_i^n)\subset V_i$. Let $V'=V_1\cap V_2$. Then $C_i^j\mod V'$ is determined uniquely, hence we have $p^{n-2k}$ variants for each $C_1^j$, and $C_2$ is reconstructed from $C_1$. Thus we have $p^{n(n-2k)}$ pairs.

In total, we get $N\cdot p^{n(n-2k)}$ pairs. In fact, this is an upper bound, since the rank of $C_i$ may be less than $p-k$, and in this case one pair will correspond to several pairs $(V_1,V_2)$.

You may easily obtain a bound for $N$, but it would be better to know the relation between $p$ and $n$...

If $A$ is degenerate, then this bound should increase, since we just need $V_1+V_2$ to contain the space generated by the rows of $A$.