Here, $\mathbb Z{_{15}}_{P}$ is the localization of the integers by the prime ideal $P$, that is, $\mathbb Z{_{15}}_{P}$=$D^{-1}Z{_{15}}$, for $D = \mathbb Z{_{15}}-P$.
I tried this problem by brute force, first by guessing that $P = (3)$. Then I tried computing the elements $a/b$. For example, I found that $1/2 = 8/16=8, 5/2 = 40/16=40/1=10, 7/2=56/16=11.$
But then I think I ran into a problem, because this is more than 3 integers. Any suggestions? And are there better ways to solve this than using brute force?
A prime ideal $P$ is also maximal, given that your rings are finite. Therefore modding out by $P$ inverts everyone that isn't in $P$. Localizing at $P$ does exactly the same thing : it inverts everyone not in $P$.
But the latter is the universal way to do so, which means there is a unique morphism $R_P\to R/P$ that is such that the composite $R\to R_P\to R/P$ is the canonical projection. Then you have to see what the map $R_P\to R/P$ looks like : it is clearly surjective, because $R\to R/P$ is. How about injectivity ? Who dies in the process ?
Well, precisely the elements of $P$, right ? Or more precisely, the images of the elements of $P$ under $R\to R_P$. So to get injectivity we only need to prove that if $x\in P$, then $x=0$ in $R_P$. That's where the specific ring structure of $\mathbb{Z}/15$ comes into play.
There are only two prime ideals : $(3)$ and $(5)$. In either case, for any $x\in P$, there is $y\notin P$ such that $xy=0$. This implies that $x=0$ in $R_P$.
So in conclusion, $R_P\to R/P$ is an isomorphism. This works more generally for any $\mathbb{Z}/n$ where $n$ is squarefree; the essential property being : if $x\in P$ there exists $y\notin P$ such that $xy=0$ (and "$P$ is a maximal ideal")