= Solution
Take $s=x-x^\ddagger$ with $0<x^\ddagger<a$. A trajectory crossing to the right at the barrier top goes downhill on exit and cannot return in this model. Its initial Boltzmann weight contains $e^{-\beta V_0}$; negative-velocity trajectories are on the reactant side at the commitment time. Therefore
$$
C_{fs}^{\rm top}=\frac{e^{-\beta V_0}}{2\pi\hbar}\int_0^\infty\frac p m e^{-\beta p^2/(2m)}dp
=\boxed{\frac{e^{-\beta V_0}}{2\pi\hbar\beta}}.
$$
Dividing by the same $Q_r$ gives the result in the energy-integral calculation. No width appears in the classical rate because every classically allowed trajectory passes through the flat barrier.
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