Prove that if $AB = 0$, then rank(A) + rank(B) ≤ p Let $A$ be an $m \times$ n matrix and $B$ be an $n \times p$ matrix.
I understand that since $AB=0$, the column space of $B$ is contained within the nullspace of $A$. Does this mean that $\operatorname{rank}(B) \leq  \operatorname{nullity}(A)$?
How do I proceed to show that $\operatorname{rank}(A)  + \operatorname{rank}(B) \leq p$ ?

Yes, you may indeed deduce that the rank of $B$ is less than or equal to the nullity of $A$.
From there, simply apply the rank-nullity theorem (AKA dimension theorem).

Counterexample to question as stated:
$$
A = \pmatrix{0&1&0\\0&0&1\\0&0&0}
,\quad
B = \pmatrix{1\\0\\0}
$$
$B$ is $3 \times 1$ and $AB = 0$, but $\operatorname{rank}(A) + \operatorname{rank}(B) = 3 > 1$.
