Walkthrough: 1) Pa,Pb,Pc,Pd are the Humpty points so $H_aP_aBH_cP_cD$ and $H_bP_bAH_dP_dC$ are concyclic. 2)Also $M_{ac}P_d.M_{ac}B=M_{ac}D.M_{ac}P_b=M_{ac}C^2$. 3)This means $P_d,B,P_b,D$ are concyclic. Now applying Radical Center on $P_dBP_bD$,$ABCD$,$H_bP_bAH_dP_dC$ we get $AC,BD,P_bP_d$ concur at a point. Similarly $P_aP_c$ also pass through that point of concurrency. 4)Now applying PoP gives the required..

Cool! . We begin with an easy Lemma. $\textbf{LEMMA: }$Triangle $ABC$ has orthocenter $H$. Then $\measuredangle BHC = \measuredangle CAB$. $\textbf{PROOF: }$It is well known that reflection $H'$ of $H$ over $BC\in \odot(ABC) \implies \measuredangle BHC=\measuredangle CH'B=\measuredangle CAB$. We focus on $\triangle ADC$ first. $\textbf{CLAIM: }P_B$ is the $D-$Humpty Point of $\triangle ADC$. $\textbf{PROOF: }$Note that $DM_{ac}$ is the $D$ median of $\triangle ADC$ and also, as $P_B$ is the projection of $H_B$ onto $DM_{ac}$, we get $H_BP_B \perp P_BD$. So merging together the facts $P_B\in DM_{ac}$ and $H_BP_B \perp P_BD$, we conclude that $P_B$ is the $D-$Humpty Point of $\triangle ADC$. So using our symmetry, we conclude the following. $\textbf{CLAIM: }P_B$ is the $D-$Humpty Point of $\triangle ADC$. $\textbf{CLAIM: }P_C$ is the $A-$Humpty Point of $\triangle BAD$. $\textbf{CLAIM: }P_D$ is the $B-$Humpty Point of $\triangle CBA$. $\textbf{CLAIM: }P_A$ is the $C-$Humpty Point of $\triangle DCB$. Now from the properties of HM point, we also have that $P_B \in \odot(AH_BC)$. Now using our Lemma we get $\measuredangle AH_BC=\measuredangle CDA=\measuredangle CBA=\measuredangle AH_DC\implies AH_BCH_D$ is cyclic. This along with the facts that $P_B \in \odot(AH_BC)$ and $P_D \in \odot(AH_DC) \implies AP_BH_BCP_DH_D$ is cyclic. Similarly $BP_AH_ADP_CH_C$ is cyclic. Thus we conclude the following. $\textbf{CLAIM: }AP_BH_BCP_DH_D$ and $BP_AH_ADP_CH_C$ are both cyclic. Now again from the properties of HM point, we know that $\odot(P_BDA)$ is tangent to $AC \implies M_{ac}A^2=M_{ac}D\cdot M_{ac}P_B$. Similarly, we have $\odot(P_DBA)$ is tangent to $AC \implies M_{ac}A^2=M_{ac}B\cdot M_{ac}P_D$. So, $$M_{ac}B\cdot M_{ac}P_D=M_{ac}D\cdot M_{ac}P_B=M_{ac}A^2\implies P_BDP_DB \text{ is cyclic.}$$Similary $P_CAP_AC$ is also cyclic. So we conclude the following too. $\textbf{CLAIM: }P_BDP_DB$ and $P_CAP_AC$ are cyclic. Now by applying Radical Axis Theorem on $\{\odot(ABCD),\odot(P_BDP_DB),\odot(P_BAP_DC)\}$ we get that $\{BD,P_BP_D,AC\}$ concur at a point. Similarly by applying Radical Axis Theorem on $\{\odot(ABCD\},\odot(P_CAP_AC),\odot(P_ADP_CB)\}$ we get that $\{BD,P_AP_C,AC\}$ concur at a point. So, let $X=P_AP_C\cap P_BP_D\cap AC\cap BD$. To finish we have, \begin{align*} XP_A\cdot XP_C &= \operatorname{Pow}_{\odot P_ADP_CB}(X)\\ &= XB\cdot XD\\ &= \operatorname{Pow}_{\odot ABCD}(X)\\ &= XA\cdot XC\\ &= \operatorname{Pow}_{\odot P_DAP_BC}(X)\\ &= XP_B\cdot XP_D\\ &\implies P_AP_BP_CP_D \text{ is cyclic.} \end{align*}

Here is a solution avoiding Humpty points, though in its main idea it feels the same as the above: Our main goal will be to prove that lines $AC, BD, P_aP_c$ and $P_bP_d$ have a common point. To prove this, we will use the following lemmas: Lemma 1. $H_a, H_c, B$ and $D$ are concyclic. (and similarly so are $H_b, H_d, A$ and $C$) Proof: \[\angle DH_cB = 180^{\circ} - \angle DAB = \angle DCB = 180^{\circ} - \angle DH_aB \Rightarrow\]\[\angle DH_cB+\angle DH_aB = 180^{\circ}\]and this lemma is proved. Lemma 2. $P_c, H_c, B$ and $D$ are conyclic (and similarly so are $P_a, H_a, B$ and $D$, also $P_b, H_b, A$ and $C$ and $P_d, H_d, A$ and $C$). Proof: Let $P_c'$ be the point on $AM_{bd}$ such that $\angle DP_c'M_{bd} = \angle ADM_{bd}$. Then triangles $DP_c'M_{bd}$ and $ADM_{bd}$ are similar and so \[M_{bd}B^2=M_{bd}D^2=M_{bd}P_c'\cdot M_{bd}A\]so triangles $BP_c'M_{bd}$ and $ABM_{bd}$ are similar (becuase if the abive relation between sides and the fact the two triangles have a common angle $\angle AM_{bd}B$). So from here $\angle M_{bd}P_c'B=\angle DBA$ so \[\angle BP_c'D=\angle ADB+\angle ABD = 180^{\circ} - \angle BAD = \angle BH_cD\]so $P_c', B, D$ and $H_c$ are concyclic. From here $\angle ADB+\angle H_cP_c'M_{bd} = \angle DP_c'M_{bd}+\angle H_cP_c'M_{bd} = \angle H_cP_c'D = 180^{\circ} - \angle H_cBD = 180^{\circ} - (90^{\circ} -\angle ADB) = 90^{\circ} + \angle ADB$ so \[\angle H_cP_c'M_{bd}=90^{\circ}\]Because there is a single point on $AM_{bd}$ with such a property, $P_c'\equiv P_c$. Because we already proved $P_c', H_c, B$ and $D$ are conyclic, our lemma is proved. Note that from the above two lemmas, because there is a single circle passing through $H_c, B$ and $D$ it follows that $H_a, H_c, P_a, P_c, B$ and $D$ are concylic. Similarly, so are $H_b, H_d, P_b, P_d, A$ and $C$. Lemma 3. $A,C,P_a$ and $P_c$ are conyclic (and similarly so are $B,D,P_b,P_d$). Proof: In the proof of the above lemma we also proved the fact $M_{bd}P_c\cdot M_{bd}A = M_{bd}B^2$ and so similarly $M_{bd}P_a\cdot M_{bd}C = M_{bd}B^2$ and so $A,C,P_a$ and $P_c$ are conyclic because $M_{bd}P_c\cdot M_{bd}A = M_{bd}P_a\cdot M_{bd}C$. Now we can finally prove our main claim: From the Radical axis theorem applied for the circumcircles of $ABCD$, $ACP_aP_c$ and $H_aH_cP_aP_cBD$ it follows that $BD, AC$ and $P_cP_a$ have a common point. Similarly, $BD, AC$ and $P_bP_d$ have a common point. So the 4 lines have a common point - let it be $I$. Then from power of point $I$ with respect to the above circles it follows that \[IP_a\cdot IP_c = IA\cdot IC = IB\cdot ID = IP_b\cdot IP_d\]so $P_a, P_b, P_c$ and $P_d$ are conyclic. $\Box$

A bunch of PoP spam: Note that $P_B$ is the second intersection of $(DH_B)$ and the reflection of $(ABCD)$ over $AC$, so $AP_DCP_B$ is clearly cyclic, and $M_{ac}P_B \cdot M_{ac}D = M_{ac}P_D \cdot M_{ac}B = M_{ac}C^2 \implies P_DBP_BD$ is cyclic. By Radical Center on $(ABCD), (AP_DCP_B), (P_DBP_BD)$, $AB \cap CD \in P_BP_D$. By Radical Center on $(ABCD), (P_DBP_BD), (P_ACP_CA)$, $AB \cap CD$ lies on the radical axis of the latter two circles, so $P_BP_D \in P_AP_C$ is on this radical axis, and so by PoP at this point, $P_AP_BP_CP_D$ is cyclic.