$\varphi'\in Hom_ R(M, N)$ differs from $\varphi$ by an element of $P Hom_ R( M, N)$, then $\varphi'$ is an epimorphism. Suppose $M$ and $N$ are finitely generated modules over a Noetherian local ring $(R,P)$ whose completions $\hat M$ (that is the inverse limit of $M/P^iM$) and $\hat N$, are isomorphic over $\hat R$(the inverse limit of $R/P^i$).
Let $\varphi \in Hom_\hat R(\hat M,\hat N)$ be an isomorphism. Show that if $\varphi'\in Hom_\hat R(\hat M,\hat N)$ differs from $\varphi$ by an element of $\hat P Hom_\hat R(\hat M,\hat N)$, then $\varphi'$ is an epimorphism.
I think I could prove this using the property of completions and Nakayama's lemma, but I wonder if this holds for $M$ and $N$ are finitely generated modules over a Noetherian local ring $(R,P)$ i.e. Removing all the hats in the claim above. Since I intened to ignore those hats as it looked easier, but I had no clues how to prove it.
 A: Consider the following form of Nakayama's lemma:

Nakayama's Lemma: Let $R$ be a commutative ring with Jacobson radical $J$ and let $M$ a finitely generated $R$-module.  If the images of $m_1, \ldots, m_n \in M$ generate $M/JM$, then $m_1, \ldots, m_n$ generate $M$ as an $R$-module, too.

An immediate consequence of this formulation is the following:

Corollary to NAK: Let $R$ be a commutative ring with Jacobson radical $J$.  Let $M, N$ be $R$-modules with $N$ finitely generated and let $\phi \in \operatorname{Hom}_R (M, N)$.  If $\bar{\phi}: M/JM \rightarrow N/JM$ is surjective then $\phi$ is surjective.

Note that we haven't used anything about Noetherianity yet.  We just need that $N$ is finitely generated.  If $R$ has a unique maximal ideal $\mathfrak{m}$, then the Jacobson radical of $R$ is equal to $\mathfrak{m}$.
In your problem statement, we have $\phi, \phi' \in \operatorname{Hom}_R(M, N)$ with $N$ finitely generated, $\phi$ an isomorphism, and $\phi - \phi' \in \mathfrak{m} \operatorname{Hom}_R(M, N)$.  Consequently, $\bar{\phi}$ and $\bar{\phi'}$ agree as homomorphisms $M/\mathfrak{m}M \rightarrow N/\mathfrak{m}N$, and Nakayama's lemma implies that $\phi'$ is already surjective.
In fact we can go further.  Here's another corollary to Nakayama's lemma:

Another Corollary to NAK: Let $R$ be a commutative ring, $M$ a finitely generated $R$-module, and $\phi: M \rightarrow M$ surjective.  Then $\phi$ is injective, too.

The trick to prove this is to consider $M$ as an $R[x]$-module by letting $x$ act on $M$ as $\phi$.  Now by construction we have $xM = M$.  Nakayama's lemma in its most common form now guarantees the existence of an element $f \in R[x]$ such that $f-1 \in xR[x]$ (i.e. $f_0 = 1$) and $fM = 0$.  If $\phi(m) = 0$ then the property $f_0 = 1$ implies that $f(m) = m$, while the property $fM = 0$ implies that $f(m) = 0$.  Hence $m = 0$, and $f$ is injective.
Now we can put this together to see that the composition $\phi^{-1} \phi': M \rightarrow N \rightarrow M$ is surjective.  Hence it is injective by our second corollary to NAK, and hence $\phi'$ must be injective.  So $\phi'$ is actually an isomorphism!! We summarize below:

Summary Let $R$ be a commutative ring with Jacobson radical $J$.  Let $M, N$ be $R$-modules with $N$ finitely generated and let $\phi, \phi' \in \operatorname{Hom}(M, N)$ such that $(\phi - \phi')(M) \subseteq JN$.  If $\phi$ is an isomorphism then $\phi'$ is an isomorphism, too.

The Noetherian and local assumptions are really only relevant to your problem in its original form, where you want to talk about completions, where you want to ensure that $\hat{N}$ remains finitely generated.
