= Solution
Interpret the requested measurement as distinguishing all four stated orthogonal <eigenstates>, hence as their rank-one <projective measurement>. If they have indistinguishable degenerate <eigenvalues>, this assumption need not hold; the identity observable, for example, cannot reveal the basis and gives no contradiction.
Let Bob initially prepare $|0\rangle_B$. Alice encodes a bit by preparing $|0\rangle_A$ or $|1\rangle_A$, without changing Bob's initial <reduced density matrix>. If Alice prepares zero, the global input is the first <eigenstate>, so nondemolition leaves Bob in $|0\rangle$ with certainty. If Alice prepares one, the rank-one <Lüders rule>, after discarding the outcome, dephases Bob in the rotated basis
$$
|u\rangle=\cos\theta|0\rangle+\sin\theta|1\rangle,
\qquad |v\rangle=\sin\theta|0\rangle-\cos\theta|1\rangle.
$$
His output is $\rho_B=\cos^2\theta|u\rangle\langle u|+\sin^2\theta|v\rangle\langle v|$. A computational-basis measurement then gives
$$
\boxed{\Pr_B(1\mid A=0)=0,\qquad
\Pr_B(1\mid A=1)=2\sin^2\theta\cos^2\theta=\tfrac12\sin^2(2\theta).}
$$
It is positive for every $0<\theta\leq\pi/4$. Repeating the experiment would transmit Alice's choice across a spacelike interval, violating <quantum no-signalling>. The <controlled-basis measurement causality obstruction> therefore excludes the entire nonzero interval, including $\pi/4$.
At \b[$\theta=0$], the basis is the computational product basis, up to an irrelevant sign on the fourth vector. Alice and Bob each measure $Z$ locally, then compare their results later. Each product <eigenstate> is preserved. Thus being a product basis is insufficient for an instantaneous <quantum nondemolition measurement>: a remote party's choice of which local basis is measured can still cause signalling.
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