= Solution
If $p>1/2$, complementing every received bit converts the channel into a binary symmetric channel with crossover probability $1-p<1/2$. This deterministic and invertible output relabelling does not change the <channel capacity>. Since $h_2(p)=h_2(1-p)$,
$$
\boxed{C(p)=1-h_2(p)=C(1-p).}
$$
\b[Thus the capacity rises again from zero as $p$ increases beyond $1/2$, reaching one at $p=1$, where every bit is flipped deterministically and can be recovered by complementation.]
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