= Solution
Use the same protocol: distribute one resource photon through $\mathcal C$, and send Alice's two classical bits error-free in the $D,A$ alphabet from part a. At $p=1/2$ the distributed resource is
$$
\rho_{23}=\tfrac12|\Phi^+\rangle\langle\Phi^+|
+\tfrac12|\Psi^+\rangle\langle\Psi^+|,
\qquad |\Psi^+\rangle=(I\otimes X)|\Phi^+\rangle.
$$
For the second component Bob has an extra $X$ before his conditional correction. Conjugation by that correction changes $X$ at most by a sign, which cancels in the <density operator>. This <bit-flipped Bell resource teleportation> therefore induces the bit-flip map,
$$
\rho_{\rm tel}=\tfrac12|\psi\rangle\langle\psi|+
\tfrac12X|\psi\rangle\langle\psi|X.
$$
Using the squared pure-target <quantum fidelity> convention,
$$
\boxed{F_\psi=\langle\psi|\rho_{\rm tel}|\psi\rangle
=\frac12\left(1+|\langle\psi|X|\psi\rangle|^2\right)
=\frac12\left(1+4[\operatorname{Re}(\alpha^*\beta)]^2\right).}
$$
The maximum is $1$ precisely for the <Pauli X eigenstates> $D,A$. The minimum is $1/2$ whenever $\operatorname{Re}(\alpha^*\beta)=0$, the great circle with zero $x$ component of the <Bloch vector>, including $H,V$ and both circular <polarizations>. Uniform averaging over pure <qubits> gives $\overline F=2/3$.
The resource-distribution and classical-encoding assumptions are essential. If a perfect shared pair had already been supplied independently and this channel were used only for the robustly encoded classical bits, teleportation would remain perfect even at $p=1/2$. Conversely, unprotected $H,V$ transmission of Alice's correction bits would add classical errors. The stated expression applies to the resource sent through the given channel and the error-free classical encoding explicitly available in part a.
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