= Solution
Conjugation by a <Pauli operator> preserves its matching <Bloch vector> component and reverses the other two. Summing all three conjugations therefore sends $\mathbf s$ to $-\mathbf s$. The specified <quantum depolarizing channel> has retention factor
$$
\eta=p-\frac{1-p}{3}=\frac{4p-1}{3},\qquad
\mathbf s\longmapsto\eta\mathbf s.
$$
This <Pauli-mixture parametrization of qubit depolarization> differs from a convention in which $p$ itself is the retention factor. For $0\leq p\leq1$, $-1/3\leq\eta\leq1$. A pure input gives output eigenvalues $(1\pm|\eta|)/2$, and hence entropy $h_2((1+\eta)/2)$ by symmetry of <binary entropy>. Mixed inputs have a smaller output radius and at least this much entropy. Every average output has entropy at most one. Thus
$$
\chi(\mathcal E_{\rm out})\leq1-h_2\left(\frac{1+\eta}{2}\right).
$$
Two equiprobable orthogonal pure inputs have antipodal output <Bloch vectors>, maximally mixed average output, and this same minimal individual output entropy. They attain the bound. Applying the coding theorem gives
$$
\boxed{C_{\rm prod}(\Lambda)=1-h_2\left(\frac{2p+1}{3}\right)
\quad\text{bits per channel use}.}
$$
At $p=1$ it is one; at $p=1/4$ it is zero; at $p=0$ it is $1-h_2(1/3)\simeq0.0817$. Negative retention reverses the labels of the two signals but does not eliminate their distinguishability.
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