= Solution
Write a qubit density operator as $\rho=(I+\mathbf r\cdot\boldsymbol\sigma)/2$, with <Bloch vector> $|\mathbf r|\leq1$. The <depolarizing channel> sends $\mathbf r$ to $p\mathbf r$. Its output eigenvalues are $(1\pm |p||\mathbf r|)/2$, so its <Von Neumann entropy> is minimized at $|\mathbf r|=1$. Define
$$
\boxed{q=(1-p)/2}.
$$
The minimum output entropy is $h(q)$, since binary entropy is symmetric under $q\mapsto1-q$. The usual probabilistic-mixture convention has $0\leq p\leq1$. The argument also holds throughout the physical qubit CPTP range $-1/3\leq p\leq1$, where this $q$ still lies in $[0,1]$.
To handle multiple channel uses, let a classical message $x$ select a product input $\rho_x=\bigotimes_{j=1}^n\rho_{xj}$ with probability $\pi_x$. Memorylessness gives product outputs $\sigma_x=\bigotimes_j\Phi(\rho_{xj})$. Additivity of <Von Neumann entropy> on tensor products gives $S(\sigma_x)\geq n h(q)$. The average output $\bar\sigma=\sum_x\pi_x\sigma_x$ acts on $n$ qubits, so $S(\bar\sigma)\leq\log_2(2^n)=n$. Thus the <Holevo quantity> obeys the <product-input block bound for a qubit depolarizing channel>
$$
\chi=S(\bar\sigma)-\sum_x\pi_xS(\sigma_x)\leq n[1-h(q)].
$$
The <Holevo bound> says the classical <mutual information> obtained by any measurement, including a joint measurement on all outputs, is at most $\chi$.
For a reliable code with $M$ equiprobable messages and decoding error $P_e$, <Fano's inequality> gives $H(X\mid\hat X)\leq h(P_e)+P_e\log_2(M-1)$. Consequently $(1-P_e)\log_2M\leq n[1-h(q)]+h(P_e)$. Letting the error tend to zero proves \b[$\boxed{\mathcal I(\Phi)\leq1-h((1-p)/2)}$.]
The bound is tight: equal use of $|0\rangle,|1\rangle$ and computational-basis output measurements produce a <binary symmetric channel> with crossover probability $q$. Ordinary classical coding achieves $1-h(q)$ bits per use.
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