= Solution
The <posterior predictive probability> averages the conditional probability over the <Bayesian posterior>. Using its <gamma distribution> density,
$$
p_0=\frac{B^A}{\Gamma(A)}\int_0^\infty\lambda^{A-1}e^{-(B+1)\lambda}\,d\lambda
=\boxed{\left(\frac{b+T}{b+T+1}\right)^{a+n}}.
$$
For fixed $a,b$ and $n/T\to\widehat\lambda<\infty$,
$$
\log p_0=-A\log(1+1/B)=-\frac{A}{B}+O(A/B^2)=-\frac nT+o(1).
$$
Consequently $p_0/e^{-n/T}\to1$, and both tend to $e^{-\widehat\lambda}$. This is reasonable because the <posterior mean> tends to the observed rate and the <posterior variance> tends to zero. Averaging $e^{-\lambda}$ then approaches evaluating it at the estimated rate: \b[with abundant observations, predictive uncertainty about the rate vanishes].
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