= Solution
Average the <conditional probability> in (i) over $Y\sim q=\widetilde q/Z_q$:
$$
\begin{aligned}
\mathbb P(\mathrm{accept})
&=\int_\Omega\frac{\widetilde q(y)}{Z_q}
\frac{\widetilde p(y)}{M\widetilde q(y)}\,dy\\
&=\frac1{MZ_q}\int_\Omega\widetilde p(y)\,dy
=\boxed{\frac{Z_p}{MZ_q}}.
\end{aligned}
$$
This is exactly the <acceptance mass for an unnormalized rejection envelope>. In normalized form the envelope is $p\leq(MZ_q/Z_p)q$, giving the same acceptance <probability>. Here $M$ bounds the ratio of the unnormalized functions, so its value need not equal the envelope constant for the normalized <probability density functions>.
The squeeze changes the number of expensive target evaluations, while preserving the per-proposal acceptance <probability>. The expected number of proposals per accepted draw is $MZ_q/Z_p$.
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