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First adopt the uniform-angular-motion approximation implicit in the requested formula. Set the line of sight to longitude zero, the $b$ exoplanet transit to $t=0$, and the next $c$ exoplanet transit to $t=\tau$, with $0\le \tau <P_c$. Write $n_b=2\pi /P_b$, $n_c=2\pi /P_c$, with $P_c>P_b$. The next conjunction satisfiesThe common longitude is $n_{b}t$ modulo $2\pi$. Subtracting one $2\pi$ gives the conjunction longitude from a transit time lagSubstitution of the period ratio yieldsIt is exact in this uniform-angle model; no first-order expansion in $\delta$ is necessary here. For other choices of which exoplanet transit is used, choose the appropriate whole-turn branch.

For the finite-eccentricity orbit specified earlier, this is an approximation. Let $f_L$ and $M_L$ be the true anomaly and mean anomaly at the line of sight. The eccentric planet has $M_c(t)=M_L+n_c(t-\tau )$ and true angular advance $n_c(t-\tau )+{\Delta }_e(t)$, whereTrue conjunction therefore requires $(n_b-n_c)t+n_c\tau -{\Delta }_e(t)=2\pi$. The PDF omits this term, which is generally $O(e_c)$.

For a concrete counterexample take $p=q=1$, $\delta =0$, $e_c=1/5$, $\tau =P_b/4$, and a line of sight along $c$'s periapsis. The uniform model predicts $t=7P_b/4$ and longitude $3\pi /2$. At that time $M_c=3\pi /2$, but Kepler's equation gives $f_c\ne 3\pi /2$, so the planets are not truly aligned. Accurate finite-eccentricity conjunctions must be found from the corrected equation.