Let $A$ be a Noetherian ring, consider $A[x]$, let $Q$ be a maximal ideal in $A[x]$, does hold that for the prime ideal $P:= A \cap Q$
$$((A/P)[x])_{\bar{Q}} \simeq ((A/P)_P[x])_{\bar{Q}}$$
where $\bar{Q} := Q \mod PA[x]$. Does a formal proof involve universal property? Is there any intuition behind it? At first glance $(A/P)_P$ seems (actually is) bigger, hence if the isomorphism holds it should be thanks to the localization in $\bar{Q}$.
Any help or proof would be appreciated.