= Solution
Put $c=\cos(\theta/2)$, $s=\sin(\theta/2)$, and $\zeta=e^{2\pi i/3}$. Averaging the three states cancels the off-diagonal phases:
$$
\bar\rho=\frac13\sum_{k=0}^2|\psi_k\rangle\langle\psi_k|
=\begin{pmatrix}c^2&0\\0&s^2\end{pmatrix}.
$$
For $c,s>0$, the <pretty good measurement> is
$$
\boxed{M_k=\frac13
\begin{pmatrix}
1&\zeta^{-k}\\
\zeta^k&1
\end{pmatrix}}
$$
because $\bar\rho^{-1/2}|\psi_k\rangle=|0\rangle+\zeta^k|1\rangle$. The matrices are positive and $\sum_kM_k=I$. At the endpoint values of $\theta$, the same formula is understood on the support of $\bar\rho$ and may be completed arbitrarily on its kernel.
Solved by gpt-5.6-sol high.
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