Someone else asked a question that seems related to this one here:
Explanation why $R_P=(S^{-1}R)_{P(S^{-1}R)}$
However, I'm wrestling with the notation, so I think I'll understand this problem better if someone is willing to answer it directly. Given a commutative ring $R$ with 1, $P$ a prime ideal and $S = R - P$, I need to show $S^{-1}RP$ is the unique maximal ideal in $S^{-1}R$. I could try to show, equivalently, that $S^{-1}R/ S^{-1}RP$ is a field, but I'm not sure if that's going to be helpful here.
$S^{-1}R = R_P$ means that everything not in $P$ gets an inverse. The set $PR_P$ is the set of fractions where the numerator is in $P$ (and the denominator is not in $P$ since that's the definition of $R_P$). That is,
$$ PR_P = \left\{ \frac{a}{b} : a \in P, b \notin P \right\} $$
The elements not in $PR_P$ are of the form $a/b$ where both $a$ and $b$ are not in $P$.
If I have such a fraction then I can take the reciprocal $b/a$ since $a \notin P$. What this is telling me is that the elements of $R_P$ which are not in $PR_P$ are already invertible in $R_P$. Hence localizing at $PR_P$ does nothing:
$$ R_P = (R_P)_{PR_P}. $$
But more importantly, if $PR_P$ is a proper ideal such that elements not in $PR_P$ are invertible then $PR_P$ must be maximal since to increase $PR_P$ would be to add a unit and we would no longer have a proper ideal.