 Solution

For $0\le q<1$, use repeat-until-success coding over an erasure channel: delete every erasure symbol from the received sequence and retain the non-erased bits in order. Each message bit contributes a string of erasures followed by exactly one correct non-erased copy. The channel encoder advances to the next bit only after this successful copy, so the retained bits are precisely $W_1,\dots ,W_n$. HenceIndeed, $\mathrm{Pr}(M_j>k)=q^k\to 0$, so each stage and the finite collection of $n$ stages terminate almost surely. The protocol uses channel feedback: Alice must know whether an erasure occurred before deciding whether to repeat. This feedback is implicit in the stated algorithm and is not counted as a forward channel use.

At $q=1$, no stage ever finishes, so there is no terminating zero-error decoder for arbitrary messages. This endpoint is excluded from the operational claim above.