= Solution
Assume $p<1/2$, and put $a=\theta/p$ and $r=p/q\in(0,1)$. The <detailed balance equations> give
$$
\frac{\pi_{k+1}}{\pi_k}=\frac{p}{q}\frac{k+a}{k+1}\frac{k+1+\theta}{k+\theta}.
$$
Multiplying from $0$ to $k-1$, the last factor telescopes, while the product involving $a$ is evaluated by the <Gamma function>:
$$
\pi_k=\pi_0\frac{k+\theta}{\theta}\frac{\Gamma(k+a)}{\Gamma(a)k!}r^k\quad(k\geq0).
$$
Consequently the requested <invariant distribution> can be expressed using the paper's power series as
$$
\boxed{\pi_k=\frac{(k+\theta)\Gamma(k+\theta/p)}{k!\,g_{p,\theta}(p/q)}\left(\frac pq\right)^k,\quad k\geq0.}
$$
The assumed convergence of $g_{p,\theta}(r)$ ensures that this normalization is finite and positive.
For an explicit normalizing constant, the <binomial series> gives
$$
H(r)=\sum_{k=0}^\infty\frac{\Gamma(k+a)}{k!}r^k=\Gamma(a)(1-r)^{-a}.
$$
The <termwise differentiation of a power series> is valid for $|r|<1$, so
$$
\begin{aligned}
g_{p,\theta}(r)&=rH'(r)+\theta H(r)\\
&=\Gamma(a)(1-r)^{-a-1}\bigl[ar+\theta(1-r)\bigr]\\
&=2\theta\Gamma(a)(1-r)^{-a-1}.
\end{aligned}
$$
The last equality uses $r=p/q$ and $a=\theta/p$: $ar+\theta(1-r)=\theta/q+\theta(q-p)/q=2\theta$. Therefore
$$
\boxed{\pi_k=\frac{(1-r)^{a+1}}{2\theta\Gamma(a)}\frac{(k+\theta)\Gamma(k+a)}{k!}r^k,\qquad \pi_0=\frac12(1-r)^{a+1}.}
$$
These positive terms sum to one and satisfy the <detailed balance equations>, so part (a)(ii) verifies stationarity. Irreducibility makes the <invariant distribution> unique.
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