= Solution
At exact conjunction the planet's force is radial and has no instantaneous <torque>, but the approach and departure do not cancel when conjunction is displaced from <apoapsis>. Write the conjunction longitude as $\lambda_c=\varpi_d+\pi+\delta$. At conjunction,
$$
\phi=\lambda_c-\varpi_d=\pi+\delta,
\qquad
\sin\phi=-\sin\delta.
$$
The resonant part of the given evolution law has $\dot a_d=-8e_da_dD\sin\phi$. Thus a conjunction after apoapsis, $\delta>0$, gives $\sin\phi<0$ and transfers <angular momentum> to the dust, increasing $a_d$. A conjunction before apoapsis removes angular momentum, while one exactly at apoapsis has zero secular transfer by symmetry.
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