 Solution

Put $q={\cos }^2(\pi /8)=(2+\sqrt{2})/4$ and $1-q={\sin }^2(\pi /8)$. In the computational basis, input $|a⟩$ has outcome probabilities $(q,1-q)$, while $|b⟩$ has $(1-q,q)$. The marginal outcomes are equally likely; the Bayesian posterior probabilities areGuess $a$ after zero and $b$ after one. Each guess is correct with probability $q\simeq 0.853553$ and wrong with probability $1-q\simeq 0.146447$. Neither outcome is certain evidence for one input.

For a real state $|{\psi }_{\theta }⟩=\cos \theta |0⟩+\sin \theta |1⟩$, the Born rule in the Hadamard basis gives $P(\pm \mid \theta )=(1\pm \sin 2\theta )/2$. Both angles $\pi /8$ and $3\pi /8$ have $\sin 2\theta =1/\sqrt{2}$. Thus both preparations give $P(+)=q$ and $P(-)=1-q$. By Bayes' theorem, either outcome leaves the two input hypotheses equally likely. The biased outcome frequencies carry no information about which state was prepared; the best success probability with this measurement is $1/2$.