 Solution

A fixed proposal-count budget does not bias the accepted values. For each trial, acceptance has probability $a$, and conditional on acceptance its label has law $p$ independently of the other trials. Conditional on the full accept/reject pattern among the first $N^{\ast }$ proposals, the successful labels are independent with common law $p$. Therefore $K\sim \mathrm{Binomial}(N^{\ast },a)$, and conditional on $K=k$, the $k$ retained samples are independent draws from $p$. When $K=0$ there is simply no sample. This concerns a budget fixed independently of the proposal values; it does not assert that an empirical average is defined on an empty output.