 Solution

Quantum no-signalling says that, without communicating a measurement outcome, any local trace-preserving operation performed on subsystem $A$ leaves all outcome probabilities for measurements on subsystem $B$ unchanged; equivalently, it leaves the reduced density matrix ${\rho }_B$ unchanged.

The two Bell states are orthogonal. Applying the inverse Bell-state preparation circuit, namely a Controlled-NOT gate from $A$ to $B$ followed by a Hadamard gate on $A$, givesA computational-basis measurement therefore distinguishes them perfectly, soIf only $B$ is accessible, however,Every measurement on $B$ consequently has the same outcome distribution under both hypotheses. With equal prior probabilities no information is available, andthe error probability of a random guess.