Let the hexagon be inscribed in the unit circle on the complex plane. With small letters we denote coordinates of considered points. Well known theorem If points $X,Y,Z,T$ lie on the unit circle on complex plane and lines $XY,ZT$ aren't parallel, then their point of intersection is $$\frac{xy(z+t)-zt(x+y)}{xy-zt}$$ Now we can see that $$m=\frac{ad(k+l)-kl(a+d)}{ad-kl}$$$$p=\frac{ac(k+l)-kl(a+c)}{ac-kl}$$$$q=\frac{bd(k+l)-kl(b+d)}{bd-kl}$$$$n=\frac{bc(k+l)-kl(b+c)}{bc-kl}$$Thus $$n-l=k\cdot\frac{(l-b)(l-c)}{bc-kl}$$$$k-p=-l\cdot\frac{(k-a)(k-c)}{ac-kl}$$$$m-q=kl\cdot\frac{(a-b)(d-k)(d-l)}{(ad-kl)(bd-kl)}$$$$k-m=-l\cdot\frac{(k-a)(k-d)}{ad-kl}$$$$l-q=-k\cdot\frac{(l-b)(l-d)}{bd-kl}$$$$n-p=kl\cdot\frac{(b-a)(c-k)(c-l)}{(bc-kl)(ac-kl)}$$Because $|k|=|l|=1$, $\forall_{z\in \mathbb C}\ |z|=|-z|$ and $\forall_{x\in \mathbb C}\forall_{z\in \mathbb C}\ |x|\cdot |y|=|xy|$ we have $$NL \cdot KP \cdot MQ =\left|\frac{(l-b)(l-c)(k-a)(k-c)(a-b)(d-k)(d-l)}{(bc-kl)(ac-kl)(ad-kl)(bd-kl)}\right| =KM \cdot PN \cdot LQ$$

Of course there is also trigonometric solution. Let $O$ be the center of the circle with radius length $R$ and denote $2\alpha_1=\angle AOB,...,2\alpha_6=\angle KOA$. Then $\alpha_1+...+\alpha_6=180^\circ$. Law of sines in triangle $DKM$: $$KM=2R\cdot\sin\alpha_5\cdot\frac{\sin\alpha_6}{\sin(\alpha_3+\alpha_4+\alpha_6)}$$Law of sines in triangle $NBL$: $$NL=2R\cdot\sin\alpha_2\cdot\frac{\sin\alpha_3}{\sin(\alpha_1+\alpha_3+\alpha_6)}$$Law of sines in triangle $BQL$: $$LQ=2R\cdot\sin\alpha_2\cdot\frac{\sin(\alpha_3+\alpha_4)}{\sin(\alpha_2+\alpha_5)}$$Law of sines in triangle $AKP$: $$KP=2R\cdot\sin\alpha_6\cdot\frac{\sin(\alpha_4+\alpha_5)}{\sin(\alpha_3+\alpha_6)}$$Law of sines in triangle $CPN$ and $CPL$: $$PN=\frac{CP\cdot\sin\alpha_1}{\sin(\alpha_2+\alpha_4+\alpha_5)}=2R\cdot\sin\alpha_1\sin\alpha_3\cdot\frac{\sin(\alpha_4+\alpha_5)}{\sin(\alpha_2+\alpha_4+\alpha_5)\cdot\sin(\alpha_3+\alpha_6)}$$Law of sines in triangle $MDQ$ and $KDQ$: $$MQ=\frac{DQ\cdot\sin\alpha_1}{\sin(\alpha_3+\alpha_4+\alpha_6)}=2R\cdot\sin\alpha_1\sin\alpha_5\cdot\frac{\sin(\alpha_3+\alpha_4)}{\sin(\alpha_3+\alpha_4+\alpha_6)\cdot\sin(\alpha_2+\alpha_5)}$$Now we are able to check that required equality is true indeed.

We want to show that $$S = \frac{KM}{MQ} \cdot \frac{QL}{LN} \cdot \frac{NP}{PK} =1$$Note that $\frac{KM}{MQ} = \frac{KD}{DQ} \cdot \frac{sin(KDM)}{sin(QDM)} = \frac{KD}{DQ} \cdot \frac{KA}{AB}$. Similarly, $\frac{LQ}{LN} = \frac{BQ}{BN} \cdot \frac{LD}{LC}$ and $\frac{PN}{PK}=\frac{CN}{CK} \cdot \frac{AB}{AK}$ Multiplying these, we get $$S = \frac{KD \cdot CN \cdot BQ \cdot DL}{DQ\cdot CK\cdot BN \cdot CL}$$Because $ABLCDK$ is cyclic, we see that $\frac{KD}{DQ}= \frac{BL}{LQ}$ and $\frac{CN}{CK} = \frac{LN}{BL}$. So , $$S = \frac{LN \cdot BQ \cdot DL}{LQ \cdot BN \cdot LC}$$. Also, $\frac{DL}{LQ}= \frac{BK}{BQ}$, so $S=\frac{BK\cdot LN}{BN\cdot LC}$ and $\frac{BK}{BN}=\frac{CL}{LN}$ $$S= \frac{CL \cdot LN}{LN \cdot CL}=1$$hence we're done

This problem is a direct consequence of the Haruki lemma