 Solution

Assume an anti-degradable channel transmits perfectly with encoder $\mathcal{E}$ and decoder $\mathcal{D}$. Apply its Stinespring isometry to $\mathcal{E}(\rho )$, producing receiver system $Q$ and environment $E$. The receiver obtains $\rho$ by $\mathcal{D}$; anti-degradability lets the environment simulate the receiver output using $\mathcal{N}$, and then $\mathcal{D}\circ \mathcal{N}$ also produces $\rho$.

Apply these two local decoding channels simultaneously to $Q$ and $E$. Each marginal of the resulting bipartite state is the original pure state $\rho$. A bipartite state with a pure marginal is a product, so the joint output is $\rho \otimes \rho$. We have therefore constructed one quantum channel mapping every pure $\rho$ to $\rho \otimes \rho$, contradicting the no-cloning theorem for two pure states. Hence no anti-degradable channel can transmit arbitrary quantum information perfectly in one use.