= Solution
Prepare the <three-vertex graph-state wire>: start all three <qubits> in $|+\rangle$ and apply <Controlled-Z gates> on edges $1-2$ and $2-3$. In this paper's <equatorial qubit measurement> convention the basis vectors can be chosen as
$$
|v_r(\theta)\rangle=\frac{|0\rangle+(-1)^re^{i\theta}|1\rangle}{\sqrt2}.
$$
The given identity means that measuring an input qubit at angle $\theta$, with outcome $r$, teleports its state to the next qubit as $X^rU(\theta)|\psi\rangle$. Each outcome has probability $1/2$, since $X^rU(\theta)$ is unitary. This is <one-bit teleportation> with a known <Pauli frame>.
Measure vertex 1 at angle $\alpha$ and record $r$. Then measure vertex 2 at angle $\gamma=(-1)^r\beta$, taken modulo $2\pi$, and record $s$. The controlled-Z on edge $2-3$ commutes with the first measurement, so the two teleportation steps can be applied successively. The remaining normalized state is
$$
X^sU(\gamma)X^rU(\alpha)|+\rangle.
$$
For $r=0$ this is $X^sU(\beta)U(\alpha)|+\rangle$. For $r=1$, the relation proved above gives $U(-\beta)X=e^{i\beta}ZU(\beta)$. Thus in every branch the remaining state is
$$
e^{ir\beta}X^sZ^rU(\beta)U(\alpha)|+\rangle.
$$
The scalar is a <global phase> and has no effect on probabilities.
Finally measure vertex 3 in the <computational basis>, obtaining $z$, and report
$$
\boxed{k=z\oplus s.}
$$
The $Z^r$ byproduct changes only computational-basis phases; $X^s$ changes the output label by $s$. Hence the corrected bit has exactly the ideal circuit's <probability distribution> in every branch, not merely on average. The procedure uses only single-qubit measurements after preparing the graph state, adaptive classical angle selection, and deterministic classical output processing.
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