If $R$ is an integral domain with field of fractions $K$ and $A$ is a fractional ideal of $R$ in $K$ (i.e., an $R$-submodule of $K$ such that $dA\subset R$ for some nonzero $d\in R$), then define $A'=\{x\in K\mid xA\subset R\}$. For any prime ideal $P$ in $R$ the localization $(A')_{P}$ of $A'$ at $P$ is a fractional ideal $R_{P}$ in $K$.
$\mathbf{Problem}$. If $A$ is finitely generated, then $(A')_{p}=(A_{P})'$.
I have no idea how to use the fact that $A$ is finitely generated. I know that localization commutes with finite products, but $A'$ seems too complicated to write out what it looks like in general.
Let me explain how the finite generation is used to show that $(A_P)'\subset (A')_P$ and leave you to check the other direction.
If $x\in (A_P)'$, the $x\in K$ and $xA_P\subset R_P$. If $a_1,\ldots, a_n$ generate $A$, we get $xa_i\in R_P$ for all $i$ and thus, you can find $s\not\in P$ such that $sxa_i\in R$ for all $i$ (notice the use of finite generation). Thus, $sx\in A'$ which implies $x\in (A')_P$.