For each object $c$ in $\mathcal{C}$, let $c^* : [\mathcal{C}, \mathcal{A}] \to \mathcal{A}$ be evaluation at $c$. It is an exact functor, so if a left adjoint $c_! : \mathcal{A} \to [\mathcal{C}, \mathcal{A}]$ exists, $c_!$ will preserve projective objects. Assume $\mathcal{C}$ has $\le \alpha$ morphisms and $\mathcal{A}$ has coproducts for families of $\le \alpha$ objects. Then the left adjoint $c_!$ exists and can be computed the following formula:
$$(c_! A) (c') = \mathcal{C} (c, c') \odot A$$
(Here, $X \odot A$ denotes the coproduct of $X$-many copies of $A$.)

Now, let $F$ be an object in $[\mathcal{C}, \mathcal{A}]$. For each object $c$ in $\mathcal{C}$, choose a projective cover $P_c \twoheadrightarrow F c$. By adjointness, we obtain morphisms $c_! P_c \to F$ in $\mathcal{A}$; note that the composite $P_c \to c^* c_! P_c \to F c$ is the epimorphism we started with, so $c^* c_! P_c \to F$ is an epimorphism in particular. Now, form the object $P = \bigoplus_{c \in \operatorname{ob} \mathcal{C}} c_! P_c$; then there is a morphism $P \to F$ such that the components $P c \to F c$ are all epimorphisms. Furthermore, each $c_! P_c$ is projective, and the class of projective objects is closed under coproducts, so we have obtained the required projective cover of $F$.

I'm afraid I do not have a reference, but the above is essentially the same as the proof that (say) $[\mathcal{C}, \mathbf{Ab}]$ has enough projectives. It can be further generalised to the case where $\mathcal{C}$ is preadditive and $[\mathcal{C}, \mathcal{A}]$ is the category of additive functors.

finite dimensional$k$-vector spaces, then $\mathcal{A}$ has enough projective for the cardinality of $Ob(\mathcal{C})=\{\ast\}$ but there are no finite dimensional $k[x]$-modules. $\endgroup$2more comments