After many diagrams by hand and computer I do not see the collinearity mentioned.. Please check the wording very carefully!

The wording seems all correct to me. Tough one to proof.. Regards David

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Since $EBCF$ is cyclic we get $\triangle EBG\sim \triangle EFC$ $\Longrightarrow$ $\tfrac{EB}{EG}=\tfrac{EF}{EC}..(\star)$, so from $AB\parallel CD$ we get $EA.ED$ $=$ $EB.EC...(\star\star)$ $\Longrightarrow$ combining $(\star)$ and $(\star\star)$ and $\measuredangle DEF=\measuredangle GEA$ we get $\triangle EAG\sim \triangle EFD$ $\Longrightarrow$ $\measuredangle EGA$ $=$ $\measuredangle EDF$ $=$ $\beta-\theta$ and $\measuredangle EAG$ $=$ $\measuredangle EFD$ $=$ $\theta$ and let $\measuredangle EFB=\alpha$ $\Longrightarrow$ by angle-chasing we get $\measuredangle DBF=\measuredangle AGB=\alpha+\theta...(\blacksquare)$ $\Longrightarrow$ by $(\blacksquare)$ we get $\odot (GAD)$ $\cap$ $\odot (BFD)$ $\in$ $AD$ hence $G\in \odot (AHBMN)$ $\Longrightarrow$ by Pascal theorem in $(BGANMH)$ we get $D=AH\cap BN$, $Q=GN\cap AM$ and $P$ $=$ $GH$ $\cap$ $BM$ are collinear hence $P$, $Q$ and $D$ are collinear.

FabrizioFelen wrote: Since $EBCF$ is cyclic we get $\triangle EBG\sim \triangle EFC$ $\Longrightarrow$ $\tfrac{EB}{EG}=\tfrac{EF}{EC}..(\star)$, so from $AB\parallel CD$ we get $EA.ED$ $=$ $EB.EC...(\star\star)$ $\Longrightarrow$ combining $(\star)$ and $(\star\star)$ and $\measuredangle DEF=\measuredangle GEA$ we get $\triangle EAG\sim \triangle EFD$ $\Longrightarrow$ $\measuredangle EGA$ $=$ $\measuredangle EDF$ $=$ $\beta-\theta$ and $\measuredangle EAG$ $=$ $\measuredangle EFD$ $=$ $\theta$ and let $\measuredangle EFB=\alpha$ $\Longrightarrow$ by angle-chasing we get $\measuredangle DBF=\measuredangle AGB=\alpha+\theta...(\blacksquare)$ $\Longrightarrow$ by $(\blacksquare)$ we get $\odot (GAD)$ $\cap$ $\odot (BFD)$ $\in$ $AD$ hence $G\in \odot (AHBMN)$ $\Longrightarrow$ by Pascal theorem in $(BGANMH)$ we get $D=AH\cap BN$, $Q=GN\cap AM$ and $P$ $=$ $GH$ $\cap$ $BM$ are collinear hence $P$, $Q$ and $D$ are collinear. Yes, this is also the official solution.

Hello. My approach: I will use the following lemma: Lemma:Given is a triangle $\triangle{ABC}$ and a point $D\in BC$.Then $\frac{\sin \hat{BAD}}{\sin \hat{CAD}}=\frac{BD}{CD}\cdot \frac{AC}{AB}$ Proof:See here. We will next prove that the points $C,D,H,G$ are concyclic.Consider $T\equiv FH\cap BC$ and $S\equiv FG\cap DA$. Obviously $\overline{FGES}$ is the angle bisector of $\angle{BEC}$. We have $\angle{SHT}=\angle{SHF}=\angle{DBF}=\angle{EBC}+\angle{CBF}=\angle{EBC}+\frac{\angle{BEG}}{2}=\angle{EGC}$,thus $SHTG$ is cyclic. Hence,it suffices to show that $ST\parallel AB$,because then $\angle{SHG}=\angle{STG}=180^{\circ}-\angle{GCD}\Rightarrow CDHG$ cyclic,q.e.d. Notice that $\angle{TFC}=\angle{EFC}=\angle{BFC}-\angle{BFH}=180^{\circ}-\angle{BEC}-\angle{BDH}=180^{\circ}-\angle{DEA}-\angle{EDA}=\angle{DAE}$. The lemma gives $\frac{\sin \hat{BFT}}{\sin \hat{TFC}}=\frac{BT}{CT}\cdot \frac{CF}{BF}=\frac{BT}{CT}\Rightarrow \frac{BT}{CT}=\frac{\sin \hat{EDA}}{\sin \hat{DAE}}=\frac{AE}{ED}$. From the angle bisector theorem we get $\frac{AE}{ED}=\frac{AS}{SD}$,thus,finally $\frac{BT}{CT}=\frac{AS}{SD}\Rightarrow ST\parallel AB$,and we are done. From the above,we also get that $AHBG$ is cyclic (simple angle-chasing),thus $A,H,B,G,M,N$ are concyclic. From Menelaus' theorem,it suffices to show that $\frac{DE}{DB}\cdot \frac{BP}{MP}\cdot \frac{MQ}{EQ}=1$. $\bullet $ From the lemma,we have $\frac{\sin \hat{MNG}}{\sin \hat{BNG}}=\frac{MQ}{EQ}\cdot \frac{EN}{MN}\Rightarrow \frac{MQ}{EQ}=\frac{\sin \hat{MBG}}{\sin \hat{BMG}}\cdot \frac{MN}{EN}=\frac{MG}{BG}\cdot \frac{MN}{EN}$. $\bullet$ Similarly,$\frac{BP}{MP}=\frac{\sin \hat{BGP}}{\sin \hat{PGM}}\cdot \frac{BG}{GM}=\frac{\sin \hat{ADC}}{\sin \hat{DAC}}\cdot \frac{BG}{GM}=\frac{AC}{CD}\cdot \frac{BG}{GM}$. Thus,it suffices to show that $\frac{DE}{DB}\cdot \frac{MN}{NE}\cdot \frac{AC}{CD}=1$.We have $\angle{EMN}=\angle{ABD}=\angle{EDC}\Rightarrow MNDC$ cyclic. Thus $\frac{MN}{CD}=\frac{EN}{EC}$,and the above relation becomes $\frac{DE}{DB}\cdot \frac{EN}{CE}\cdot \frac{AC}{EN}=1\Leftrightarrow \frac{DE}{DB}=\frac{CE}{AC}$,which is obvious,and we are done. 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I think this is a really hard problem and requires an effective solution. Here is motivation for it. First notice that we are asked to prove collinearity. We can use Menelaus, radical axis, completing angles $180$ or "Pascal Theorem" Secondly let's focus on the points which are asked to be proved and think on the relationships between them. We realize $AH \cap BN = \{D\}$, $AM \cap NG =\{Q\}$ and $HG \cap BM = \{P\}$. So it makes sense to use Pascal Theorem. Then we should prove that $AHBGMN$ is cyclic. It is already given that $AHBMN$ is cyclic. Because of that we must try to prove $G$ is on the $(AHBMN)$ also! Thanks, Burak