$P$ is projective implies $P℘$ is a free $R℘$-module

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Statement: Let $R$ be a commutative Noetherian ring and let $P$ be a finitely generated $R$-module, then $P$ is projective if and only if $P℘$ is a free $R℘$-module for all $℘$ in $Spec(R)$.

For the converse part, I have used for $F$ free module any exact sequence $0→M→N→F→0$ splits and the definition of projective modules $P$ as

Every exact sequence $0 → N → M → P → 0$, of $R$-modules and $R$-linear maps splits.

But for $(\implies way)$ I am unable to approach,