National Math Olympiad in Slovenia

Denote the circumcircle of $\triangle MBL$ by $K_1$ and the circumcircle of $\triangle MCN$ by $K_2$. The line $NM$ is tangent to $K_1$, so $\angle NML = \angle MBL$ (the tangent-chord angle theorem). Similarly, $LM$ is tangent to $K_2$, so $\angle NML = \angle NCM$. Hence,
$$\angle NCM = \angle NML = \angle MBL \qquad (3)$$
$$\angle DCB = \angle CBA = \angle NML \qquad (4)$$
We have $\angle LMB = \angle(LM, MC) = \angle MNC$ and (3), so the triangles $\triangle MBL$ and $\triangle NCM$ are similar. This implies
$$\frac{|LB|}{|MC|} = \frac{|LM|}{|MN|} = \frac{|BM|}{|CN|}. \qquad (5)$$
Since
$$\frac{|AL|}{|LB|} = \frac{|BM|}{|MC|} = \frac{|CN|}{|ND|},$$
we have
$$\frac{|AL|}{|AB|} = \frac{|BM|}{|BC|} = \frac{|CN|}{|CD|}.$$
Thus
$$\frac{|AL|}{|BM|} = \frac{|AB|}{|BC|} \qquad (6)$$
$$\frac{|BM|}{|CN|} = \frac{|BC|}{|CD|} \qquad (7)$$
From (4), (5), (6) and (7) we conclude that $\triangle ABC$, $\triangle LMN$ and $\triangle BCD$ are similar. Now, denote $\angle BDC = \alpha$, $\angle CBD = \beta$ and $\angle DCB = \gamma$. Then
$$\begin{aligned} \angle DPA &= \pi - \angle CPD \\ &= \pi - (\pi - \angle PDC - \angle DCP) \\ &= \pi - (\pi - \angle BDC - \angle DCA) \\ &= \pi - (\pi - \angle BDC - (\angle DCB - \angle ACB)) \\ &= \pi - (\pi - \alpha - (\gamma - \alpha)) \\ &= \pi - (\pi - \alpha - \gamma + \alpha) \\ &= \pi - (\pi - \gamma) \\ &= \gamma \\ &= \angle DCB = \angle NML. \end{aligned}$$