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 $△CNP\sim △TNB$ and therefore $(TL+NL)PN=CN\cdot BN$, whence $TL=\frac{NL\cdot KP}{PN}$. Analogously, $△DQM\sim △TQB$ implies $TL=\frac{LQ\cdot KM}{MQ}$ and we have the desired result.