= Solution
Under a correctly specified <Poisson regression> with adequate expected counts, the residual <Poisson deviance> is approximately <chi-squared distribution> with $n-p=107-5=102$ <residual degrees of freedom>. The observed $D=319.86$ gives
$$
\boxed{\Pr(\chi^2_{102}\geq319.86)\approx2.62\times10^{-24}.}
$$
Thus the ordinary Poisson model gives a very poor fit under this goodness-of-fit approximation. This is consistent with <overdispersion>, although a wrong mean model or dependence can also cause lack of fit; quasi-Poisson variance inflation alone does not establish that the mean model is adequate.
For <Quasi-Poisson regression>, $\operatorname{Var}(Y_i)=\phi\mu_i$. The <Pearson dispersion estimator> is
$$
\boxed{\widehat\phi=\frac1{102}\sum_{i=1}^{107}\frac{(Y_i-\widehat\mu_i)^2}{\widehat\mu_i}.}
$$
It uses squared <Pearson residuals>, not the residual deviance divided by 102. The fitted coefficients are unchanged, and their estimated <covariance matrix> is multiplied by $\widehat\phi$. With $\widehat\phi=3.24$, standard errors increase by $1.8$, so the main treatment coefficient has
$$
t=\frac{-0.3592068}{1.8(0.1117255)}\approx-1.7862,\qquad p=2\Pr(t_{102}\geq1.7862)\approx0.0770.
$$
\b[The main Progabide coefficient is not significant at 5% after this dispersion adjustment.] R uses a <Student's t-distribution> when dispersion is estimated, as described in https://stat.ethz.ch/R-manual/R-devel/library/stats/html/summary.glm.html[its GLM summary documentation]; the large-sample normal approximation also gives the same decision.
Because an <interaction term> is present, this tests the treatment contrast at baseline zero. An overall absence of treatment influence requires the joint null $\beta_T=\beta_{TB}=0$. The printed marginal standard errors do not give their covariance, so they are insufficient to calculate that joint <Wald test>, or the treatment contrast test at an arbitrary nonzero baseline.
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