Akhil's answer works proposes a model structure, but it doesn't have the property required by the original question, i.e. that free functors are cofibrant. Also, proving factorization is much more involved than Akhil's answer would lead you to believe (in fact, this is where all the hard work is done). In this answer I'll fill in the details of that proof and at the bottom address the non-cofibrancy of free functors.

The proof that the model structure exists mimics Hirschhorn's proof that the projective model structure exists, and that proof takes about 6 pages. I'll summarize it here, but those who know it can just skip the next section where I fill in the details in Akhil's setting (i.e. with fibrations and weak equivalences defined based on $M\subset K$). The last section brings it back to the acyclic models theorem.

## Projective Model Structure

The reason factorization works in the standard projective model structure is that you transfer the model structure from the cofibrantly generated $C^{K^{disc}} = \prod_{ob(K)} C$ along a Quillen adjunction $(F,U)$, and you apply the small object argument. This transfer principle is Hirschhorn's Theorem 11.3.2. Let $I$ and $J$ be the generating cofibrations and generating trivial cofibrations for $C$. For every $k\in K$ let $I_k= I \times \prod_{k'\neq k} id_{\phi}$, where $\phi$ is the initial object of $C$. This is how to view $I$ in just one component of $C^{K^{disc}}$. The generators for $C^{K^{disc}}$ are $I_{ob(K)} = \cup_{k\in K} I_k$ and similarly $J_{ob(K)} = \cup_{k\in K} J_k$.

The existence of the projective model structure (and the fact that it's cofibrantly generated) is Hirschhorn's Theorem 11.6.1 and the proof verifies that the hypotheses of 11.3.2 hold, namely:

- $F(I_{ob(K)})$ and $F(J_{ob(K)})$ permit the small object argument.
- All maps in $U(F(J_{ob(K)})-cell)$ are weak equivalences.

For Hirschhorn, the functor $F$ takes a discrete diagram $X$ to a diagram $\coprod_{k \in K} F^k_{X(k)}$ in $C^K$ where $F^k_A = A \otimes K(k,-)$. The functor $U$ is the forgetful functor. Hirschhorn's proof of (2) requires Lemma 11.5.32, which analyzes pushouts of $F_A^k \to F_B^k$, since for any $g:A\to B$, $F(g\otimes \prod_{k'\neq k} id_\phi)$ is the map $F_A^k\to F_B^k$. The point of Lemma 11.5.32 is that if $X\to Y$ is a pushout of this map then $X(k)\to Y(k)$ is a pushouts in $C$ of a giant coproduct of copies of $g$. This coproduct is a trivial cofibration, so pushouts of it are trivial cofibrations and hence $X\to Y$ is a weak equivalence as desired.

Lemma 11.5.32 gets used again (along with the Yoneda lemma and basic facts about smallness) to prove (1). The point is that the domains of the generators are $F_A^k$ and $A$ is small relative to $I$-cof, so one can use adjointness to pass between maps in $C^K$ and maps in $C$ (i.e. one component of $C^{K^{disc}}$) and use smallness there to get the desired commutativity of colimit and hom.

## Modified Projective Model Structure

To prove Akhil's proposed model structure has factorization you need to define the correct functor $F$ AND you have to verify (1) and (2). It turns out that Akhil has the correct $F$ (if you soup it up so it goes from $C^{K^{disc}}\to C^K$ rather than $C\to C^K$), but verifying (1) and (2) are non-trivial because this $F$ doesn't automatically have Lemma 11.5.32. Anyway, the correct $F$ takes $X$ to $\coprod_{m\in M}F^m_{X(m)}$ in $C^K$. Note that it doesn't land in $C^M$ because $K(m,-)\neq M(m,-)$. This is good, since if it landed in $C^M$ you'd just end up with the projective model structure on $C^M$.

To make a long story short, you can prove an analog of Lemma 11.5.32 and follow Hirschhorn to get the modified projective model structure. A better way is to notice that on the coordinates in $K - M$ all maps are trivial fibrations, the cofibrations are isomorphisms, and both the generating cofibrations and the generating trivial cofibrations are $\lbrace id_\phi \rbrace$. So if $M$ is empty you get the trivial model structure (Hovey's example 1.1.5) on $C^K$, and if $M$ is a one-point set you get a product of $C$ with a bunch of copies of that trivial model structure. Of course if $M=K$ you get the usual projective model structure.

With this observation, a better way to verify existence of this modified projective model structure is to take a product where in coordinate $k\in M$ one uses the projective model structure and in coordinate $k\not \in M$ one uses the trivial model structure. Note that this modified projective structure is done in Johnson and Yau's "On homotopy invariance for algebras over colored PROPs" for the case where $K$ is a group acting on $C$ and $M$ is a subgroup. Unfortunately, I only found this reference after working the above out for myself.

## Connection to Acyclic Models Theorem

The main reason why I don't think this model structure helps with the acyclic models theorem is what I alluded to in my comment to Akhil's answer. You really need $K$ to be small, and this rules out all the interesting examples to which the acyclic models theorem applies. The other reason is the observation above about getting the trivial model structure in coordinates $k\notin M$. With this, you can't have the "free functors being cofibrant" that the OP asks for. Recall that a free functor here means there are models $M_\alpha \in M$ and $m_\alpha \in F(M_\alpha)$ such that for all $X\in K$ the set $\lbrace F(f)(m_\alpha)\rbrace$ is a basis for $F(X)$. Because cofibrations are isomorphisms in the coordinates $k\notin M$, the only cofibrant object there is the initial object, i.e. the empty functor. These free functors are not of that form because their behavior is not restricted in that way on $k\notin M$.