= Solution
Let $f_j(z)=f(p^jz)$ for $0\le j\le r$, and write $a=a_p(f)$ and $W_Mf=\varepsilon f$. The <Hecke operator> $U_p$ acts on the Fourier expansion by $\sum a_nq^n\mapsto\sum a_{pn}q^n$. Since $p\mid M$, the primitive relation is $a_{pn}=a_pa_n$. Hence $U_pf_0=af_0$ and $U_pf_j=f_{j-1}$ for $j\ge1$. Substituting the argument directly in the <Fricke involution> gives
$$
W_N f_j=\varepsilon p^{k(r/2-j)}f_{r-j}.
$$
Thus, in the basis $e_j=p^{jk/2}f_j$, put $b=p^{k/2}$ and let $R$ be the reversal matrix, $Re_j=e_{r-j}$. The matrices are
$$
\boxed{[U_p]=\begin{pmatrix}
a&b&0&\cdots&0\\
0&0&b&\cdots&0\\
\vdots&&\ddots&\ddots&\vdots\\
0&\cdots&0&0&b\\
0&\cdots&0&0&0
\end{pmatrix},\qquad[W_N]=\varepsilon R.}
$$
If the exam's rescaling $w_N=N^{k-1}W_N$ is used, multiply the second matrix by $N^{k-1}$.
Here is an explicit irreducibility argument for the <prime-power old block at a bad prime>. Any nonzero common invariant complex subspace $V$ contains an <eigenvector> of the restricted $U_p$. Its eigenvalue is either $a$ or zero. If $a\ne0$, the $a$-eigenspace is $\mathbb Ce_0$; containing it gives $e_r$ by reversal, then every $e_j$ by successive powers of $U_p$. The kernel of $U_p$ is $\mathbb C(e_1-(b/a)e_0)$. If $V$ instead contains this vector, reversal and $U_p^r$ give
$$
U_p^rR(e_1-(b/a)e_0)=b^{r-1}(a-b^2/a)e_0.
$$
This is nonzero: the <bad-prime eigenvalue of a trivial-character newform> satisfies $a^2=p^{k-2}$ when $p\parallel M$, and $a=0$ when $p^2\mid M$, whereas $b^2=p^k$. If $a=0$, the kernel is already $\mathbb Ce_0$, and the same reversal-and-shift argument works. Therefore \b[the block has no nonzero proper subspace invariant under $U_p$ and the Fricke operator], and hence none invariant under the full ring in the question.
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