= Solution
The <stopped martingale> $Z_{n\wedge T}$ is a <martingale>. For completeness, the bounded-time stopping rule follows directly from
$$
Z_{(n+1)\wedge T}-Z_{n\wedge T}
=\mathbf1_{\{T>n\}}(Z_{n+1}-Z_n),
$$
whose <conditional expectation> given $\mathcal F_n$ is zero. Integrability at each fixed $n$ follows by expressing the stopped value as a finite sum of the integrable $Z_j$ restricted to disjoint <events>. Therefore $\mathbb E Z_{n\wedge T}=\mathbb E Z_0=1$.
The crucial bound is deterministic. Since $S_{n\wedge T}\leq b$ and $M(\tau)\geq1$,
$$
0<Z_{n\wedge T}=\frac{e^{\tau S_{n\wedge T}}}{M(\tau)^{n\wedge T}}\leq e^{\tau b}.
$$
A bounded family is <uniformly integrable>: for $K>e^{\tau b}$ all its quantities $\mathbb E[|Z_{n\wedge T}|\mathbf1_{\{|Z_{n\wedge T}|>K\}}]$ vanish. Part (c) gives $T<\infty$ <almost surely>, and then $S_T=b$. Hence
$$
Z_{n\wedge T}\longrightarrow e^{\tau b}M(\tau)^{-T}\quad\text{almost surely}.
$$
The general result for passing to expectations is the <dominated convergence theorem>: almost-sure convergence under one integrable bound implies <convergence in L1> and convergence of expectations. Applying it with the bound $e^{\tau b}$ gives
$$
1=\lim_n\mathbb E Z_{n\wedge T}=e^{\tau b}\mathbb E[M(\tau)^{-T}],
\qquad
\boxed{\mathbb E[M(\tau)^{-T}]=e^{-\tau b}.}
$$
This is the <exponential first-passage transform for an upward skip-free random walk>; it avoids an unjustified use of <optional stopping> at an unbounded <stopping time>.
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