= Solution
Keep $A=\beta GM_\star/c$ constant. A circular <circumstellar orbit> stays circular in the secular drag approximation, and $da/dt=-2A/a$ integrates to $a^2=a_p^2-4At$. The <inspiral time under Poynting–Robertson drag> to the stellar surface is
$$
\boxed{t_{\rm cs}=\frac{a_p^2-R_\star^2}{4A}\simeq\frac{a_p^2}{4A}}.
$$
The final expression treats the <star> as a point, or assumes $R_\star\ll a_p$.
For the coplanar <circumplanetary orbit> used in the preceding result, $da/dt=-3Aa/a_p^2$, so $a=a_0\exp(-3At/a_p^2)$. The <inspiral time of circumplanetary dust> to the <planet>'s surface, for $a_0>R_p$, is
$$
\boxed{t_{\rm cp}=\frac{a_p^2}{3A}\ln\frac{a_0}{R_p}},\qquad
\boxed{\frac{t_{\rm cp}}{t_{\rm cs}}=
\frac{4\ln(a_0/R_p)}{3[1-(R_\star/a_p)^2]}}.
$$
Thus the timescales have the same dependence on stellar flux and $\beta$, but planetary arrival contains a logarithm of the initial planetocentric radius. It is not always shorter: for a point <star>, $t_{\rm cp}<t_{\rm cs}$ only if $a_0/R_p<e^{3/4}$. A constant tilted-orbit average replaces three by its appropriate orientation coefficient. <Sublimation> or other removal can terminate the evolution before either idealized arrival time.
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