Let $A$ be an Artin algebra. A finite dimensional $A$-module $M$ is torsionless, that is $M$ can be embedded into a projective $A$-module. We say a map $f: M \rightarrow P$ a left $A$-approximation of $M$ if $P$ is projective and $\operatorname{Hom}(f,P'): \operatorname{Hom}(P,P') \rightarrow \operatorname{Hom}(M,P')$ is surjective for any projective module $P'$.
Then for a torsionless module $M$, how to get that it must have a left $A$-approximation?
If $\{\alpha_1,\dots,\alpha_n\}$ is a generating set for $\text{Hom}_A(X,A)$ as a module for $Z(A)$, the centre of $A$, then the map $f:X\to A^n$ given by $x\mapsto\left(\alpha_1(x),\dots,\alpha_n(x)\right)$ is a left $A$-approximation, since if $\alpha:X\to A$ is any map, then $$\alpha=z_1\alpha_1+\dots+z_n\alpha_n$$ for some $z_1,\dots,z_n\in Z(A)$, so $\alpha=g\circ f$ where $g:A^n\to A$ is given by $$g(a_1,\dots,a_n)=z_1a_1+\dots+z_na_n.$$
It’s irrelevant whether $X$ is torsionless, and the same argument works for left (or right) $\text{add}(M)$-approximations for any finitely generated $A$-module $M$. The existence of $\mathcal{C}$-approximations is only really a difficult question when $\mathcal{C}$ contains infinitely many non-isomorphic indecomposable modules.