= Solution
With equal <prior probability>[priors], the probability of a correct result from this <quantum measurement> is
$$
P_S
=\frac12|\langle b_0|a_0\rangle|^2
+\frac12|\langle b_1|a_1\rangle|^2.
$$
The two required amplitudes are
$$
\langle b_0|a_0\rangle=\cos\beta,
\qquad
\langle b_1|a_1\rangle
=\frac{\cos\beta-\sin\beta}{\sqrt2}.
$$
Therefore
$$
\begin{aligned}
P_S
&=\frac12\cos^2\beta
+\frac14(\cos\beta-\sin\beta)^2\\
&=\frac12+\frac14(\cos2\beta-\sin2\beta)\\
&=\frac12+\frac{\sqrt2}{4}
\cos\left(2\beta+\frac\pi4\right).
\end{aligned}
$$
Its maximum occurs at $\beta=-\pi/8$ and is
$$
\boxed{
P_S^{\max}
=\frac12+\frac1{2\sqrt2}
=\frac12\left(1+\frac1{\sqrt2}\right)
}.
$$
Here $|\langle a_0|a_1\rangle|=1/\sqrt2$, so $\theta=\pi/4$ and $\sin\theta=1/\sqrt2$. The stated <Helstrom-Holevo bound> is consequently exactly the same value, and the basis above is the <Helstrom measurement for the zero and plus states>.
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