 Solution

If PA is first-order consistent and already proves $\neg X$, it cannot also prove $X$. By the correctness of the ordinary arithmetized arithmetical proof relation, $\mathbb{N}\models \neg {\mathrm{Pr}}_{PA}(┌X┐)$.

If PA is inconsistent, it proves $X$ by explosion, so an actual formal proof code witnesses ${\mathrm{Pr}}_{PA}(┌X┐)$ in the standard natural numbers. ThusThe hypothesis that $X$ is refutable is essential to this first-order consistency characterization for an arbitrary fixed first-order sentence.