Prove or disprove that $NL$ is closed under polynomial many-one reductions
Let's prove the two directions separately. Suppose first that $\mathsf{NL}=\mathsf{P}$. In order to prove the statement, suppose that $B \in \mathsf{NL}$ and that there is a polytime many-one reduction from $A$ to $B$. Since $\mathsf{P}$ is closed under polytime many-one reductions and $B \in \mathsf{NL} = \mathsf{P}$, clearly $A \in \mathsf{P} = \mathsf{NL}$, proving the statement.

Suppose next that the statement is correct. Let $B$ be any non-trivial language in $\mathsf{NL}$ (for example, $B = \{\epsilon\}$), and let $A \in \mathsf{P}$ be arbitrary. Since $A \in \mathsf{P}$ there is a polytime function $f$ computing it. We can easily modify $f$ to a polytime reduction from $A$ to $B$ by fixing some YES instance and NO instance in $B$ (here we need that $B$ be non-trivial). The statement implies that $A \in \mathsf{NL}$. Since $A$ was arbitrary, we deduce that $\mathsf{P} \subseteq \mathsf{NL}$ and so $\mathsf{NL} = \mathsf{P}$.