= Solution
As in the intended independent-mark construction, take the entire sequence $(X_j)$ <independent> of the <Poisson process>, not merely independent of one fixed count. Then $Y$ is a <Compound Poisson process>. For deterministic $s<t$, its increment $Y_t-Y_s$ is <independent> of $\mathcal F_s=\sigma(Y_u:0\leq u\leq s)$. To justify this carefully, first condition on $N_s=k$. The past of $Y$ uses only the counting history up to $s$ and marks among $X_1,\ldots,X_k$. The future count has <independent> rate-$\lambda$ <Poisson process> increments, and future marks $X_{k+1},X_{k+2},\ldots$ are fresh with the common law. Their joint conditional law is the same for every $k$ and every past history, so the increment is independent even of this larger marked history, hence of $\mathcal F_s$. Marks equal to zero do not cause a problem: $\mathcal F_s$ may then reveal less than the full counting history.
By part (b), with Poisson parameter $\lambda(t-s)$,
$$
\mathbb E e^{-q(Y_t-Y_s)}
=\exp\{\lambda(t-s)[\phi(q)-1]\}.
$$
The proposed process is adapted to $\mathcal F_t$, nonnegative, and
$$
\mathbb E|Z_t|=e^{\lambda t[1-\phi(q)]}\mathbb E e^{-qY_t}=1<\infty.
$$
In particular, no finite <expectation> of a mark is required. Also $Z_s>0$, and
$$
\frac{Z_t}{Z_s}
=\exp\{-q(Y_t-Y_s)+\lambda(t-s)[1-\phi(q)]\}
$$
is <independent> of $\mathcal F_s$ and has mean one. Therefore the <exponential Laplace martingale of a compound Poisson process> satisfies
$$
\boxed{\mathbb E[Z_t\mid\mathcal F_s]
=Z_s\,\mathbb E[Z_t/Z_s]=Z_s\qquad(0\leq s<t).}
$$
This proves every condition for a <continuous-time martingale> in the requested <filtration>. When $q=0$, $\phi(0)=1$ and the same conclusion reduces to the constant <martingale> $Z_t=1$.
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