If $M$ is a $R$-module in a $K$-vector space $V$, is $\widehat M \cap V = M_{\mathfrak p}$?

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Let $R$ be a Noetherian integral domain with field of fractions $K$. Let $\mathfrak p$ be a prime ideal of $R$ and assume that the $\mathfrak p$-adic completion $\widehat R$ is a domain. Write $\widehat K = \operatorname{Frac}(\widehat R)$. Let $V$ be a finite dimensional $K$-vector space and $M$ a finitely generated $R$-submodule of $V$. We may regard $M$ as a subset of the $\mathfrak p$-adic completion $\widehat M \cong M \otimes_R \widehat R$, which we may view as a subset of $V \otimes_K \widehat K$.

Is $V \cap \widehat M = M_{\mathfrak p}$?, the intersection being taken in $V \otimes_K \widehat K$.

Clearly, the localization $M_{\mathfrak p}$ is contained in $V \cap \widehat M$. I wondered about this in an attempt to translate results about localizations to results about completions.

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It is no restriction to assume $R = R_{\mathfrak p}$. It is then true if $R$ is a discrete valuation ring (i.e. if $R$ is principal). Then $M$ is free, so in $V \otimes_K \widehat K$ we have $M \widehat R \cap M K = M (\widehat R \cap K)$, so the question reduces to proving the equality $\widehat R \cap K = R$. This is true because $R$ is the valuation ring of $K$.