= Solution
The moment generating function is $M_X(t)=\mathbb E(e^{tX})$ wherever that <expectation> is finite. A useful <continuity> theorem says that if the moment generating functions $M_n$ are finite on a common <open interval> containing zero and converge pointwise there to the moment generating function of a probability distribution, then the corresponding variables converge in distribution to that law. Equivalently one may assume a finite limiting function on such a neighborhood, with the usual <continuity> at zero, and obtain the uniquely determined limiting law.
For the gamma density, integrating the exponential against the density and changing variable to $(\lambda-t)x$ gives
$$
\boxed{M_X(t)=\left(\frac\lambda{\lambda-t}\right)^\alpha,\qquad t<\lambda.}
$$
For the waiting time to one head, <independence> of successive tosses gives $\Pr(Y=j)=p(1-p)^{j-1}$ for $j\ge1$. Summing the <geometric series> gives
$$
\boxed{M_Y(t)=\frac{pe^t}{1-(1-p)e^t},\qquad (1-p)e^t<1.}
$$
The waiting times between consecutive heads are <independent> copies of $Y$, so their sum $N$ has moment generating function $M_Y(t)^k$. Consequently
$$
M_{2pN}(t)=\left[\frac{pe^{2pt}}{1-(1-p)e^{2pt}}\right]^k\longrightarrow(1-2t)^{-k}.
$$
For fixed $t<1/2$, numerator and denominator have leading terms $p$ and $p(1-2t)$ as $p\downarrow0$. A fixed neighborhood of zero is inside the convergence domain for all sufficiently small $p$, so the <continuity> theorem applies. The limiting distribution is
$$
\boxed{\operatorname{Gamma}(\text{shape }k,\ \text{rate }1/2),}
$$
equivalently a <chi-squared distribution> with $2k$ <degrees of freedom>.
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