= Solution
Let the ideal two-<qubit> output before final <quantum measurements> be
$$
|\Omega\rangle=E_{23}J(\alpha)_2|+\rangle_2|+\rangle_3.
$$
The PDF places $J(\alpha)$ only on the upper wire, followed by the <Controlled-Z gate> and a <quantum measurement in the computational basis> on each output wire. Measure <vertex> $0$ of the square in the <computational basis>, obtaining $r$, then <vertex> $1$ of the surviving path in the fixed <equatorial qubit measurement> basis at angle $\alpha$, obtaining $s$. Put $t=r\oplus s$. The previous part gives $E_{23}X_2^tJ(\alpha)_2Z_3^r|++\rangle$.
Commute its <Pauli frame> through the <Controlled-Z gate>. Since $E_{23}X_2^t=X_2^tZ_3^tE_{23}$ and $E_{23}$ commutes with $Z_3^r$, the actual output is
$$
X_2^{r\oplus s}Z_3^s|\Omega\rangle.
$$
Finally measure <vertices> $2,3$ in the <computational basis>, with raw results $d_2,d_3$. A <Pauli X gate> flips a <computational basis> result, whereas a <Pauli Z gate> changes only its phase. Thus the purely classical correction is
$$
\boxed{k=d_2\oplus r\oplus s,\qquad l=d_3.}
$$
This <four-cycle graph-state simulation of an entangle-and-measure circuit> reproduces the full joint output distribution, not just each marginal. All measurement bases are fixed beforehand, and no physical byproduct correction is necessary. As a check, the ideal <Controlled-Z gate> is diagonal in the output basis, so $P(k,l)=\tfrac12\cos^2(\alpha/2)$ for $k=0$ and $\tfrac12\sin^2(\alpha/2)$ for $k=1$, independently of $l$. The measurement procedure gives exactly these probabilities after its classical relabelling.
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