I've learned the answer to my question, which I write here for future questioners:
First, we assume that $\Lambda$ is Artinian, so that any finitely generated $\Lambda$-module has a composition series, and also any submodule of a finitely generated module is itself finitely generated.
We first show that any finitely generated module has a unique projective cover, constructed as follows. Let $M$ be finitely generated, so there is some free, hence projective, module $P_1$ such that $f:P_1\twoheadrightarrow M$. Suppose also that $P_1$ is minimal with respect to its decomposition into indecomposable modules (that is, no summand of $P_1$ lies in the kernel of $P_1\twoheadrightarrow M$). Now, suppose that $g:P_2\twoheadrightarrow M$ is another such surjection, with $P_2$ projective and minimal. By projectivity, we must have maps $P_1\to P_2$ and $P_2\to P_1$ (commuting with the appropriate diagrams), which we now show are isomorphisms.
Suppose that $h:P_1\to P_2\to P_1$ is not an isomorphism. Then by Fitting's lemma (here is where we use the fact that $P_1$ has a composition series), there is some $n$ such that $P_1=\mathrm{im}(h^n)\oplus\ker(h^n)$, with $\ker(h^n)\ne 0$. But notice that $f\circ h^n=f$, so that $\ker(h^n)\in\ker(f)$, which contradicts the minimality hypothesis of $P_1$. It follows that $h$ is an isomorphism. Applying the same argument to $P_2\to P_1\to P_2$, we see that $P_1\cong P_2$. Define the projective cover of $M$, $P_M$, to be the unique (up to isomorphism) minimal projective module which surjects onto $M$. Let $\Omega(M)$ be the kernel of the map $P_M\twoheadrightarrow M$.
Now, I define a minimal projective resolution of $M$ to be a projective resolution $P^{\bullet}\xrightarrow{\partial} M$ such that no indecomposable summand of $P_n$ lies in the kernel of $\partial_n$ for all $n\ge 0$.
Begin constructing a projective resolution of $M$ by setting $P_0=P_M$, and $\Omega^1(M)=\Omega(M)$. Having defined $P_{n-1}$ and $\Omega^{n}(M)$, let $\Omega^{n+1}(M)=\Omega(\Omega^n(M))$, and $P_{n}=P_{\Omega^{n}(M)}$. By construction, the resolution is minimal, and because the projective cover of a finitely generated module is unique up to isomorphism, the resolution is unique up to isomorphism. Notice that at each step we use the fact that $\Omega^n(M)$ is finitely generated because it is a submodule of a finitely generated module.
Notice also we've defined $\Omega^n(M)$ differently than Benson has, but the two definitions are equivalent.
$\newcommand\ideal{\mathfrak}$This holds, in general, on every PID $A$.
Let $\ideal a$ be an ideal of $A/Aa$ and $\varepsilon:\ideal a\to A/Aa$ be an $A/Aa$-module homomorphism.
Then $\ideal a=Ab/Aa$ for some $b\in A$ and $a\in Ab$, so that $a=bc$ for some $c\in A$.
Let $y\in A$ such that $\varepsilon(b+Aa)=y+Aa$.
Then there exists $d\in A$ such that $cy=da=dbc$ that's $y=bd$.
Thus the $A$-module homomorphism
\begin{align}
&A\to A/Aa&&x\in A\mapsto dx+Aa
\end{align}
factors through a projection $A\to A/Aa$ modulo $Aa$ giving rise to an $A$-module homomorphism $A/Aa\to A/Aa$ extending $\varepsilon$.
Best Answer
Here is one such, once you notice that $R=K[x]/x^n$ is self-injective.
$$0\to K\to R\stackrel{x}{\to}R\stackrel{x^{n-1}}{\to} R\stackrel{x}{\to} R\stackrel{x^{n-1}}{\to} R\cdots$$