Let's denote the second intersection point of the line BI with the circumcircle of $\triangle ABC$ as W (fig. 35).
Clearly, W is a midpoint of the smaller arc AC, and points M, W, and N lie on the same line.

Hence, it is enough to prove that triangles $PMN$ and $IWN$ are similar (which will imply that $\angle PNM = \angle INW$ and the desired collinearity). In order to prove the similarity, we are going to show that $\frac{NM}{NW} = \frac{PM}{IW}$.
Let $\angle B = 2\beta$, then $\angle MNA = \beta$ and
$$\frac{NM}{NW} = \frac{PM}{IW} \cdot \frac{NA}{NW} = \cos^2 \beta.$$

Considering a segment PM yields that PM is a midline of a trapezoid AXYC, where X, Y are the projections of A and C onto the line $A_1C_1$ respectively. Going further, we get
$$2\mathrm{PM} = \mathrm{CY} + \mathrm{XA} = \mathrm{CA}_1 \sin \mathrm{YA}_1\mathrm{C} + \mathrm{C}_1 \sin \mathrm{XC}_1\mathrm{A} = (\mathrm{CA}_1 + \mathrm{AC}_1)\cos \beta = \mathrm{AC}\cos \beta,$$
because $\angle \mathrm{YA}_1\mathrm{C} = \angle \mathrm{XC}_1\mathrm{A} = 90^\circ - \beta$, and $\mathrm{CA}_1 + \mathrm{AC}_1 = \mathrm{AC}$, because $A_1$ and $C_1$ are the points in which inscribed circle touches the sides of a triangle $\triangle ABC$.
Finally, from the trillium theorem, $IW = CW$, meaning that
$$\frac{PM}{IW} = \frac{AC}{2CW} \cos \beta = \frac{CM}{CW} \cos \beta = \cos^2 \beta,$$
finishing the proof.