We have: (1): $AE+HP=AH+EP$ (2): $HP+DG=HD+PG$ (3): $EB+PF=EP+BF$ (4): $PF+GC=PG+FC$ Adding these above equations gives $(AB+CD)-(BC+AD)=2(HF-EG)$. So it suffices to show that $HF=EG$, which I haven't done yet.

The problem has a generalization; See the attachment below, $\ell_1,\ell_2,\ell_3,\ell_4$ are common external tangents of the referred circles (different from AB,BC,CD,DA), meeting each other at $X,Y,Z,W.$ We prove that $ABCD$ has an incircle $\Longleftrightarrow$ $XYZW$ has an incircle. Using the notations given by the diagram, we get $XY=P_1Q_1-XM-YN=AB-AP_1-BQ_1-XM-YN \ \ (1)$ $WZ=P_3Q_3-ZL-WO=DC-CP_3-DQ_3-ZL-WO \ \ (2)$ $YZ=P_2Q_2-YN-ZL=BC-BQ_1-CP_3-YN-ZL \ \ (3)$ $XW=P_4Q_4-XM-WO=AD-AP_1-DQ_3-XM-WO \ \ (4)$ Combining $(1)+ (2)$ and $(3) +(4)$ yields $XY+WZ-(YZ+XW)=(AB+DC)-(BC+AD)$ $XY+WZ=YZ+XW \Longleftrightarrow AB+DC=BC+AD$ $XYZW$ has an incircle $\Longleftrightarrow$ $ABCD$ has an incircle. P.S. See also the highlighted theorem in the topic Tangential quadrilateral.

Attachments:

I managed to complete my proof. Let the incentres of $BFPE$, $AEPH$, $HPGD$ and $PHCG$ be $K$, $L$, $M$ and $N$, respectively. Observe that $L$, $P$, $N$ are collinear, $K$, $P$, $M$ are collinear, and that these lines are perpendicular. Note that $\angle KEL=\angle LHM=\angle MGN=\angle NFK=90$ and that $\angle EPL=\angle HPL=\angle FPN=\angle GPN$. Let these angles equal $x$. Now set up a Cartesian Coordinates system where $LPN$ is the y-axis, $KPM$ is the x-axis and $P$ is the origin. Let $L=(0,a), M=(b, 0), N=(0, -c)$ and $K=(-d, 0)$. Let $H=(p, ptan(90-x))$. Using $LH$ is perpendicular to $MH$, we obtain $x=\frac{a+btan(90-x)}{(sec(90-x))^2}$ so the length of $PH$ is $\frac{a+btan(90-x)}{sec(90-x)}$. Similarly, we can determine the lengths of $PG$, $PE$ and $PF$, and then it immediately follows that $HF=EG$. The proof above is much nicer than this, but this is my ugly coordinates solution for anybody interested. Nice problem, though!

Nice angle chasing exercise!

Let $I_A$, $I_B$, $I_C$, and $I_D$ be the incenters of $HAEP$, $EBFP$, $FCGP$, and $GDHP$. Since $\angle I_API_B + \angle I_CPI_D = 180^\circ$, it follows that $P$ has an isogonal conjugate $Q$ in $I_AI_BI_CI_D$. Claim. This point $Q$ is the desired incenter. Proof. This is pure angle chasing; we show that $\overline{AQ}$ bisects $\angle A$. Using directed angles modulo $180^\circ$, we have \begin{align*} \angle DAQ + \angle BAQ & = \angle (\overline{AD}, \overline{I_AI_D}) + \angle I_DI_AQ + \angle I_BI_AQ + \angle (\overline{AB}, \overline{I_AI_B})\\ & = \angle (\overline{AD}, \overline{I_AI_D}) + \angle PI_AI_B + \angle PI_AI_D + \angle (\overline{AB}, \overline{I_AI_B})\\ & = \angle (\overline{I_AI_D}, \overline{EG}) + \angle I_DI_BI_A + \angle I_BI_DI_A + \angle (\overline{I_AI_B}, \overline{HF})\\ & = \angle (\overline{I_AI_D}, \overline{EP}) + \angle I_BI_DI_A + \angle I_DI_BI_A + \angle (\overline{I_AI_B}, \overline{HP})\\ & = \angle (\overline{I_BI_D}, \overline{EP}) + \angle (\overline{I_DI_B}, \overline{HP})\\ & = \angle (\overline{I_BP}, \overline{EP}) + \angle (\overline{I_DP}, \overline{HP})\\ & = 0 \end{align*}and we're done. It's not hard to show that $Q$ lies inside $I_AI_BI_CI_D$ and thus inside $ABCD$, so $Q$ does lie on the internal bisectors (and not the external bisectors).

Someone correct me if I'm wrong, but I believe this problem is fairly simple by taking a suitable projective transformation .

Segment chasing

Rename $E,F,G,H,P$ as $P,Q,R,S,O.$ Let the incircles of $APOS,BQOP,CROQ,$ and $DSOR$ be $\omega_1,\omega_2,\omega_3,$ and $\omega_4,$ respectively. Let $\omega_1$ and $\omega_2$ touch $\overline{AB}$ at $A_1$ and $A_2.$ Let $\omega_1$ touch $\overline{PR}$ and $\overline{QS}$ at $P_1$ and $Q_1.$ Similarly define $B_1,B_2,$ etc and $P_2,Q_2,$ etc. WLOG let $A_1$ be closer to $P$ than $A_2$ and let $A_3$ be closer to $Q$ than $A_4.$ Notice $$2PP_1+P_1P_2=A_1A_2=Q_1Q_2=OP_2+OP_1=2OP_2+P_1P_2$$so $PP_1=OP_2.$ Hence, $$PR-QS=(PA_1+A_1A_3+RA_3)-(QQ_3+Q_3Q_1+SQ_1)=(PA_1-QQ_3)+(RA_3-SQ_1)=0.$$By Pitot, $DR=OR+SD-OS$ etc, so $$AB+CD=2(OR+OP-OS-OQ)+SD+CQ+BQ+AS=BC+AD+2(PR-QS)=BC+AD.$$$\square$

Work with naming conventions from 2015 ISL G7. Claim: $ABCD$ has an incircle. Proof. Let the tangents from $\omega_1$ and $\omega_2$ meet $\overline{AB}$ at $W_1$ and $W_2$. Consider clockwise and label the tangents from $\overline{BC}$, $\overline{CD}$, $\overline{DA}$ by $X_i$, $Y_i$ and $Z_i$. Similarly denote the tangency points of $\omega_1$ and $\omega_2$ to $\overline{PR}$ by $T_1$ and $T_2$. Continuing, label the tangents at $\overline{QS}$ from $\omega_2$ and $\omega_3$ by $T_3$ and $T_4$, and so on. Then observe that, \begin{align*} AB + CD - AD - BC &= W_1W_2 + Y_1Y_2 - X_1X_2 - Z_1Z_2\\ &= \frac{1}{2} \left(2W_1W_2 + 2Y_1Y_2 - 2X_1X_2 - 2Z_1Z_2 \right) \end{align*}Note then that $W_1W_2 = T_3T_7$ due to the fact that they are both common external tangents to $\omega_1$ and $\omega_2$. Then we have, \begin{align*} \frac{1}{2} \left(2W_1W_2 + 2Y_1Y_2 - 2X_1X_2 - 2Z_1Z_2 \right) &= \frac{1}{2} \left(2T_3T_8 + 2T_4T_7 - 2T_2T_5 - 2T_6T_1 \right)\\ &= \frac{1}{2}(OT_1 + OT_2 + OT_5 + OT_6 - OT_3 - OT_4 - OT_7 - OT_8)\\ &= 0 \end{align*}whence the last step follows by equal tangents. Hence $AB + CD - AD - CB = 0 \iff AB + CD = AD + CB$ and hence $ABCD$ is tangential. $\square$