= Solution
Put $x=g_0g$ in the definition of $U(g_0)|\alpha ij\rangle$ and use
$$
M_\alpha(g_0^{-1}x)
=M_\alpha(g_0)^\dagger M_\alpha(x).
$$
One obtains
$$
\boxed{U(g_0)|\alpha ij\rangle
=\sum_{k=1}^{d_\alpha}
M_{\alpha,ki}(g_0)|\alpha kj\rangle}.
$$
Thus every $d_\alpha^2$-dimensional subspace $\mathcal H_\alpha$ is invariant; the left regular representation acts as $M_\alpha(g_0)$ on the first matrix index and as the identity on the second.
Let $P_\alpha$ project onto $\mathcal H_\alpha$. Since the subspace is invariant under every $U(g)$, $P_\alpha$ commutes with those unitaries. Also $|gH\rangle=U(g)|H\rangle$. Therefore
$$
\boxed{\Pi_H(\alpha)
=\|P_\alpha|gH\rangle\|^2
=\|P_\alpha|H\rangle\|^2},
$$
which is independent of the coset representative $g$ and depends only on $H$.
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