= Solution
Use the nonnegative representative of reduction modulo four. The four product <eigenstates> and <eigenvalues> are
$$
\begin{array}{c|rrrr}
\text{state}&|00\rangle&|01\rangle&|10\rangle&|11\rangle\\
\text{eigenvalue}&2&0&0&2
\end{array}
$$
because $-2\equiv2\pmod4$. Consequently the observable is \b[$I+Z_AZ_B$]. It has the even sector $\operatorname{span}\{|00\rangle,|11\rangle\}$ with value two and the odd sector $\operatorname{span}\{|01\rangle,|10\rangle\}$ with value zero.
The <entanglement-assisted nondemolition parity measurement> needs just the single shared $|\Phi^+\rangle$ provided in the question. Apply $\operatorname{CNOT}_{A\to d_1}$ and $\operatorname{CNOT}_{B\to d_2}$. For a computational-basis input $|a b\rangle$, the meter becomes
$$
\frac{|a b\rangle_d+|1\oplus a,1\oplus b\rangle_d}{\sqrt2}.
$$
Measuring the meters gives $u\oplus v=a\oplus b$. For an arbitrary coherent system input, its conditional <Kraus operator> is
$$
K_{uv}=\frac1{\sqrt2}P_{u\oplus v},\qquad
P_p=\frac{I+(-1)^pZ_AZ_B}{2}.
$$
It preserves every superposition within the measured sector, so this realizes the <Lüders rule> for the two degenerate <eigenvalues>. The answer is \b[two for equal meter bits, zero for unequal meter bits]. Each individual meter bit is uniform; only later comparison reveals the parity, respecting <quantum no-signalling>. This construction is independent of the defective one-pair singlet-verification request in part (a).
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