 Solution

Identification is stronger than quantum state exclusion. Here an outcome only needs to rule out its associated preparation. Use the following orthonormal basis for a basic measurementIts defining overlaps satisfy $⟨b_m|{\varphi }_m⟩=0$ for $m=0,1,2$. By the Born rule, observing $b_m$ makes preparation ${\varphi }_m$ impossible. Measure in this basis and send its outcome label $m$; the prize is guaranteed.

For completeness, the conditional probability matrix, with rows $m$ and columns the prepared-state label $j$, isEvery column sums to one and the diagonal vanishes. Each result leaves two possible preparations, so this projective state exclusion does not contradict the impossibility of perfect identification. The family consists of antidistinguishable quantum states.