= Solution
Regard $B$ as a <linear map> $\mathbb R^n\to\mathbb R^p$ and $A$ as a <linear map> $\mathbb R^p\to\mathbb R^m$ (the same argument holds over any <field>). Then
$$
\operatorname{im}(AB)=A(\operatorname{im}B)\subseteq\operatorname{im}A.
$$
Its <dimension> cannot exceed either $\dim\operatorname{im}A$ or $\dim\operatorname{im}B$, since applying a <linear map> cannot increase <dimension>. The <rank bound for a matrix product> is therefore
$$
\boxed{\operatorname{rank}(AB)\leq\min\{\operatorname{rank}A,\operatorname{rank}B\}\leq p.}
$$
The bound $p$ is attained when
$$
A=\begin{pmatrix}I_p\\0_{(m-p)\times p}\end{pmatrix},\qquad B=\begin{pmatrix}I_p&0_{p\times(n-p)}\end{pmatrix}.
$$
Their product has an $I_p$ block and zeros elsewhere, so its <rank of a matrix> is $p$. Thus the bound depending only on the given <matrix> sizes is \b[sharp].
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