= 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 $A$, gives
$$
|\phi^+\rangle\longmapsto|00\rangle,
\qquad
|\phi^-\rangle\longmapsto|10\rangle.
$$
A computational-basis measurement therefore distinguishes them perfectly, so
$$
\boxed{P_{\rm err}^{AB}=0.}
$$
If only $B$ is accessible, however,
$$
\operatorname{tr}_A|\phi^+\rangle\langle\phi^+|
=\operatorname{tr}_A|\phi^-\rangle\langle\phi^-|
=\frac I2.
$$
Every measurement on $B$ consequently has the same outcome distribution under both hypotheses. With equal prior probabilities no information is available, and
$$
\boxed{P_{\rm err}^{B}=\frac12,}
$$
the error probability of a random guess.
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