It is fairly straightforward to show that if a relation is computable/recursive, then it is representable in $PA$. That is, if $R(\vec{m})$ is a computable relation, then there is a wff $\varphi$ s.t:
$$ \begin{align*} R(\vec{m}) \text{ is true} &\implies PA \vdash \varphi(\vec{m}) \\ R(\vec{m}) \text{ is false} &\implies PA \vdash \lnot \varphi(\vec{m}) \\ \end{align*} $$
My question might be incredibly silly, but I was wondering whether the converse holds. That is, suppose I have a relation $R$ which is representable in $PA$. Does this imply that $R$ is computable?
Yes. Given an $m$, you can decide whether $R(m)$ holds by enumerating all proofs of PA until you find either a proof of $\varphi(m)$ or $\lnot \varphi(m)$. Since $\varphi$ represents $R$, you'll eventually find a proof of one or the other.