= Solution
The definition uses a fixed reference volume, so it is a <reference-volume thermal expansion coefficient>:
$$
\alpha=\frac{d}{dT}\frac{V(T)}{V_0}=\frac{dP_1}{dT}.
$$
Let $y=3e^{-\varepsilon/(k_BT)}$. Since
$$
\frac{dy}{dT}=\frac{\varepsilon}{k_BT^2}y,
\qquad \frac{d}{dy}\frac{y}{1+y}=\frac1{(1+y)^2},
$$
we obtain
$$
\boxed{\alpha(T)=\frac{\varepsilon}{k_BT^2}
\frac{3e^{-\varepsilon/(k_BT)}}{\left(1+3e^{-\varepsilon/(k_BT)}\right)^2}}.
$$
It is positive and has units of inverse temperature. Equivalently, $\alpha=[\varepsilon/(k_BT^2)]P_1(1-P_1)$: the response reflects changes in the thermally occupied excited population. The usual instantaneous fractional <thermal expansion coefficient> $V^{-1}dV/dT$ would instead equal $\alpha V_0/V$; substituting that different definition would change the answer.
\Image[/past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2023/paper-2-two-level-expansion.png]
{title=Degeneracy limits the high-temperature volume of the two-level model}
{description=Original dimensionless calculation with $t=k_BT/\varepsilon$. The plotted expansion response is $(\varepsilon/k_B)\alpha$. The high-temperature volume approaches $7V_0/4$, with vanishing response at both temperature extremes.}
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