If $Q \in (PBC)$ such that $PBQC$ is harmonic, then by angle chasing $\triangle{QBC} \sim \triangle{PAD}$, so $\frac{PB}{PC} = \frac{PA}{PD} \implies \frac{PA}{PB} = \frac{PD}{PC}$. Also, $\angle{BPC} + \angle{APD} = \angle{APB} + \angle{CPD} = 180^{\circ}$, so if $R \in (PCD)$ such that $PCRD$ is harmonic, $\triangle{PAB} \sim \triangle{RDC}$. Thus, $4 \cdot PM \cdot PK = 4 \cdot PK \cdot KR \cdot \frac{AB}{CD}$. By Power of a Point from $K$ to $(PCRD)$, $PK \cdot KR = \left(\frac{CD}{2}\right)^2$, so $4 \cdot PM \cdot PK = CD^2 \cdot \frac{AB}{CD} = AB \cdot CD$ as desired. $\square$

Take $R$ and $S$ such that $PARB$ and $PCSD$ are parallelograms, in which $P$,$M$, $R$ collinear, $P$,$K$,$S$ collinear and $P$,$N$,$Q$ collinear. [asy][asy] import geometry; size(200); pair A = (2, 5), B = (8.62, 7.62), C = (7.78, 0.71), D = (3.95, 1.56), P=(5, 4), R=(5.62,8.62), S=(6.73,-1.74), Q=(11.4,4.33); draw (P--Q, grey+dashed); pair M = midpoint(A--B); pair N = midpoint(B--C); pair K = midpoint(C--D); draw (P--N); filldraw ( A--P--B--R--cycle , opacity (1) + lightcyan , blue ) ; filldraw ( D--P--C--S--cycle , opacity (1) + lightcyan , blue ) ; draw (B--Q, green+dashed); draw (P--R, blue+dashed); draw (P--S, blue+dashed); draw (A--B--C--D--cycle); draw(arc(P, 0.5, degrees(B-P), degrees(A-P)), red); draw(arc(P, 0.5, degrees(C-P), degrees(D-P)), blue); draw(arc(B, 0.5, degrees(P-B), degrees(R-B)), blue); draw(arc(C, 0.5, degrees(P-C), degrees(S-C)), red); filldraw ( P--Q--C--cycle , opacity (0) + lightgreen , black) ; filldraw ( A--P--D--cycle , opacity (0) + lightgreen , black ) ; draw (B--C,black); dot("$A$", A, dir(180)); dot("$B$", B, dir(0)); dot("$C$", C, dir(350)); dot("$D$", D, dir(200)); dot("$P$", P, dir(110)); dot("$M$", M, dir(330)); dot("$N$", N, dir(0)); dot("$K$", K, dir(30)); dot("$R$", R, dir(90)); dot("$S$", S, dir(350)); dot("$Q$", Q, dir(30)); [/asy][/asy] Now, $\angle CPB+\angle APD=180^\circ$ so $\angle BPA+\angle DPC=180^\circ$ and as $\triangle PAD\sim \triangle CQP$ we obtain $$\frac{PA}{PD}=\frac{PB}{PC}$$ then $PARB$ and $PCSD$ are similar therefore $$\begin{align*} \frac{PR}{AB}&=\frac{CD}{PS}\Rightarrow\\ \\ AB\cdot CD&= 4PM\cdot PK \end{align*}$$ as desired

Very nice, guys! Thanks for all!

Construct parallelograms $PDXC$ and $PAYB$. Then $\angle DPA = 180^{\circ} - \angle CPN - \angle BPN$, so $\angle DPC + \angle APB = 180^{\circ}$. It follows that $\angle DXC + \angle AYB = 180^{\circ}$ so $PDXC$ and $PAYB$ are similar. From this, we find that $\frac{CD}{PY} = \frac{PX}{AB} \implies AB \cdot CD = 4PM \cdot PK$.

The main claim is that the same angle conditions also hold "on the other side", i.e. Claim: $\angle APM=\angle PDC$ and symmetric variants. Let $\angle BPN=\theta_1$ and $\angle CPN=\theta_2$. Then, by ratio lemma we have $$\frac{PC}{PB}=\frac{\sin\theta_1}{\sin\theta_2}=\frac{PD}{PA}.$$ Now, let $\angle APM=\theta_3$ and $\angle BPM=\theta_4$. Note that $\angle APD+\angle PBC=180$, so $\angle APB+\angle DPC=180$, which means that $$\angle PDC+\angle PCD=\theta_3+\theta_4.$$ Furthermore, $$\frac{\sin\angle PDC}{\sin\angle PCD}=\frac{PC}{PD}=\frac{PB}{PA}=\frac{\sin\theta_3}{\sin\theta_4},$$ and since $\angle PDC+\angle PCD=\theta_3+\theta_4$, we must have $\angle PDC=\theta_3$ and $\angle PCD=\theta_4$, as desired. The condition we want to show can be rearranged as $$\frac{BM}{PM}\times \frac{CK}{PK}=1$$ $$\frac{\sin\angle MPB\sin\angle KPC}{\sin\angle ABP\sin\angle DCP}.$$ However, as shown in our previous claim, $\angle MPB=\angle DCP$ and $\angle KPC=\angle ABP$, so we are done. Remark: There is a simpler way to finish after using the ratio lemma, which I saw from other solutions on the AoPS thread. By erecting parallelograms outside of the quadrilateral, we can directly take advantage of $\angle APB+\angle DPC=180$ and the equal ratios from the ratio lemma to induce similar triangles which proves Claim 7.1 directly. In fact, if we construct two parallelograms, one on either side, we don't even have to do the ratio lemma! Noting that the two parallelograms are similar suffices, as @dolphinday explained above. The lesson here is that when there are many midpoint conditions involved, constructing parallelograms is often useful, especially if we have length or angle conditions that "do not touch" as it lets us "move" angles and lengths.

Solved with MathLuis. Very fun problem. The great god carried - as always. Note that the given condition rewrites to, $$\begin{align*} AB \cdot CD &= 4PM \cdot PK\\ PM \cdot PK &= \frac{AB}{2} \cdot \frac{CD}{2}\\ PM \cdot PK &= BM \cdot KC\\ \frac{PM}{BM} &= \frac{KC}{KP} \end{align*}$$ which upon applying the Sine Rule on triangles $\triangle BMP$ and $\triangle PKC$, suffices to show $$\frac{\sin \angle PAB}{\sin \angle APM}=\frac{\sin \angle DPK}{\sin \angle PDC} \qquad{(1)}$$ . We now show the following key claim. Claim : $\frac{PA}{PB}=\frac{PD}{PC}$ Proof : By Ratio Lemma on $\triangle BPC$ we have that $$\frac{CP}{BP}=\frac{\sin \angle NPB}{\sin \angle CPN}=\frac{\sin \angle PAD}{\sin \angle PDA}=\frac{PD}{PA}$$ which upon rearrangement yields the desired ratio condition. Then, note that by applying the Ratio Lemma on $\triangle PDC$ and Law of Sines on $\triangle PCK$ we have $$\frac{PD}{CP}=\frac{\sin \angle CPK }{\sin \angle DPK}=\frac{CK \sin \angle PCK}{PK \sin \angle DPK}$$ Similarly we obtain that, $$\frac{PA}{PB}=\frac{\sin \angle BPM}{\sin \angle APM}= \frac{BM \sin \angle PBA}{PM \sin \angle APM}$$ Thus, $$\frac{CK \sin \angle PCK}{PK \sin \angle DPK}= \frac{BM \sin \angle PBA}{PM \sin \angle APM}$$ So, comparing this with the desired relation (1), it suffices to show $\triangle BMP \sim \triangle CKP$. This is our final claim. Claim : The triangles $\triangle BMP$ and $\triangle CPK$ are similar. Proof : Let $Q$ the Miquel point of $ABCD$. We consider the spiral similarity taking $\triangle APB \to \triangle DRC$. Using this spiral similarity centered at $Q$ we get that $$\frac{CP}{PD}=\frac{BP}{PA}=\frac{CR}{RD}$$ This along with $\angle APD=180-\angle PAD-\angle PDA=180-\angle BPN -\angle CPN=180-\angle BPC$ gives that $CRDP$ is cyclic and also an harmonic quadrilateral. Now to check all of the following chain of similarities use force overlay and the spiral similarity at $Q$, we basically have $$\triangle BMP \sim \triangle CKR \sim \triangle RDP \sim \triangle PKC$$ which was the desired claim. As per the previous reduction, this means that we are done.

We first note $\angle APB + \angle CPD = 180$. Ratio lemma tells us $$\frac{CP}{BP} = \frac{\sin \angle BPN}{\sin \angle CPN} = \frac{\sin \angle PAD}{\sin \angle PDA} = \frac{DP}{AP}.$$ Thus we can perform a spiral similarity at $P$ sending $A \mapsto D$ which makes $B'PC$ collinear. Since this transformation preserves the value of $\frac{AB \cdot CD}{PM \cdot PK}$, we are finished by noting that $PK$ and $PM'$ are midlines in $\triangle B'CD$. $\blacksquare$

quad2

@v4913 Why does the similarity between QBC and PAD implies the first equality?

@dolphinday You wrote: $\angle DPC + \angle APB = 180^{\circ}$. wouldn't that imply $\angle APD + \angle BPC = 180^{\circ}$? and hence triangle DPA isoceles? ( btw, I had the same approach as you but I couldn't carry the thought further than this T-T ) Thanks in advance!