 Solution

Let $E$ record whether the second stage erased the symbol, and write its output as $Z$. The erasure marker is outside the original output alphabet, so $E$ is determined by $Z$. Independence makes $E$ independent of both $X$ and the first output $Y$. The chain rule for mutual information therefore givesThe erased branch gives zero information because its output is constant; the unerased branch has exactly the original joint input-output law. Maximizing over the same input distributions givesThis includes $p=0$ and $p=1$. It is the capacity after independent output erasure; input-dependent erasure would not justify the independence step.