= Solution
Take observation periods to have positive lengths $t_i>0$, and let $P$ be the <Markov kernel> of a sweep which first draws $b$ given the current $\theta$ and then draws the new vector $\Theta'$. Write $A_i=a+y_i$. Conditional on $b$, <independence> and the <gamma distribution> mean and <variance> give
$$
\mathbb E\!\left[\left(\sum_i\Theta_i'\right)^2\Bigm|b\right]
=\left(\sum_i\frac{A_i}{b+t_i}\right)^2+
\sum_i\frac{A_i}{(b+t_i)^2}
\leq\left(\sum_i\frac{A_i}{t_i}\right)^2+\sum_i\frac{A_i}{t_i^2}=:D.
$$
Averaging over the <full conditional distribution> of $b$ preserves this bound, uniformly in the current state. Therefore $PV(\theta)\leq M:=1+D<\infty$. This is the mechanism in <bounded conditional moments imply a geometric drift>.
Choose $0<\rho<1$, $R>M/\rho$, and $C=\{\theta:V(\theta)\leq R\}$. Outside $C$, $PV\leq M<\rho V$; inside $C$, $PV\leq M\leq\rho V+M$. Hence
$$
\boxed{PV(\theta)\leq\rho V(\theta)+M\mathbf1_C(\theta),\qquad 0<\rho<1.}
$$
For completeness, $C$ is a <small set>, even though it approaches the boundary of the positive orthant. Choose $\delta>0$ small enough that $B=[\delta,2\delta]^n\subset C$. On $C$, $s=\sum_i\theta_i\leq\sqrt{R-1}$. The conditional <gamma distribution> of $b$ has fixed shape $na+c$ and rate $1+s$ in a compact positive interval. Its <probability density function> therefore has a positive common lower bound on $b\in[1,2]$. For those $b$, the product <probability density function> of $\Theta'$ similarly has a positive common lower bound on $B$, since every shape is positive and every rate $b+t_i$ lies in a compact positive interval. Integrating over $b\in[1,2]$ gives $P(\theta,\cdot)\geq\varepsilon\eta(\cdot)$ on $C$, where $\eta$ is the <uniform distribution> on $B$ and $\varepsilon>0$.
The transition <probability density function> is positive throughout the positive orthant, giving an <irreducible Markov chain>. Since $B\subset C$, the box minorization includes starts in the same box, which gives an <aperiodic Markov chain>. The proper <prior distributions> and bounded <Poisson distribution> likelihood have positive finite evidence, so the invariant <posterior distribution> is proper. Together with the <geometric drift condition>, these hypotheses imply <geometric ergodicity>. In particular, the argument proves the required drift rather than assuming that every <Gibbs sampler> is geometrically ergodic.
The positive-period convention matters. If zero lengths are allowed, the printed assertion can fail: take $n=1$, $a=c=1$, $t_1=y_1=0$. Then $b\mid\theta\sim\operatorname{Gamma}(2,1+\theta)$ and $\Theta'\mid b\sim\operatorname{Gamma}(1,b)$, so
$$
\mathbb E[(\Theta')^2\mid\theta]=2\mathbb E[b^{-2}\mid\theta]=\infty.
$$
This example has a proper <posterior distribution> but cannot satisfy a finite quadratic <geometric drift condition>. Thus the proof establishes the intended claim for positive observation periods, not an unrestricted extension to zero exposure.
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