Let $X, Y$ denote $P_2P_3 \cap P_{13} P_{14}$ and $P_1P_4 \cap P_{14} P_{15}$ respectively. It suffices to show that $XY || P_2 P_4.$ Let $\theta = \frac{2 \pi}{17}$. Note that the distance from $Y$ to $P_{13}P_{14}$ is $YP_{14} \cdot \sin 2 \theta$ and the distance from $Y$ to $P_2P_3$ is $P_1P_2 \sin 2 \theta$ because $P_1P_4 || P_2 P_3.$ As $\angle P_2 X P_{14} = 5 \theta,$ it suffices to show that $\frac{P_1P_2}{YP_{14}} = \frac{\sin \theta}{\sin 4 \theta.}$ To do this, the Law of Sines means that we only need to show that $P_{14} P_1 = P_{14} Y.$ This is simple angle-chasing (just show that $P_{14} Y P_1$ is isosceles with base $P_1Y$). $\square$