= Solution
Let $Y_0$ and $Y_1$ be the aggregate counts in the control and treated groups. Assuming independent physicians, their <Poisson distributions> have <means> $12\lambda$ and $12\lambda e^\beta$. Condition on $E=\{Y_0+Y_1=m\}$. By <independent Poisson conditioning>,
$$
Y_1\mid E\sim\operatorname{Bin}\left(m,p\right),\qquad
p=\frac{12\lambda e^\beta}{12\lambda+12\lambda e^\beta}
=\frac{e^\beta}{1+e^\beta}.
$$
The PDF's full table gives $Y_0=67$, $Y_1=133$ and $m=200$. Thus the <Two-group Poisson ratio conditional likelihood> is
$$
\boxed{L_c(p)=\binom{200}{133}p^{133}(1-p)^{67},\qquad0<p<1.}
$$
Conditioning the individual observations instead gives a multinomial distribution with cell probabilities $(1-p)/12$ in the first group and $p/12$ in the second; its parameter-dependent <likelihood> is the same. No factor involves $\lambda$. The conditional estimates are $\widehat p=133/200$ and $\widehat\beta=\log(133/67)\simeq0.6857$.
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