Bulgarian National Mathematical Olympiad

First solution. Denote $s = \sin \frac{\hat{A}B}{2}$, $t = \sin \frac{\hat{B}L}{2}$, $u = \sin \frac{\hat{L}C + \hat{A}K}{2}$, $v = \sin \frac{\hat{C}K}{2}$, $w = \sin \frac{\hat{D}K}{2}$ and $x = \sin \frac{\hat{L}D + \hat{A}K}{2}$. Then we consecutively have
$$\frac{NL \cdot KP \cdot MQ}{KM \cdot PN \cdot LQ} = \frac{NL}{NC} \cdot \frac{NC}{NP} \cdot \frac{KP}{AK} \cdot \frac{AK}{KM} \cdot \frac{MQ}{DQ} \cdot \frac{DQ}{LQ} = \frac{t}{v} \cdot \frac{u}{s} \cdot \frac{v}{u} \cdot \frac{x}{w} \cdot \frac{s}{x} \cdot \frac{w}{t} = 1.$$

Second solution. We choose $T \in MN$ such that $\angle NTB = \angle ADB = \angle ACB$ and $N$ is between $P$ and $T$. Then $\triangle CNP \sim \triangle TNB$ and therefore $(TL + NL)PN = CN \cdot BN$, whence $TL = \frac{NL \cdot KP}{PN}$. Analogously, $\triangle DQM \sim \triangle TQB$ implies $TL = \frac{LQ \cdot KM}{MQ}$ and we have the desired result.