Obviously $I_1,J_1,I_3,J_3$ are collinear on the internal bisector of $\angle AOB$ and $I_2,J_2,I_4,J_4$ are collinear on the internal bisector of $\angle BOC.$ In any $\triangle OAB$ we have the well-known identity $OA \cdot OB=OI_1 \cdot OJ_1$ and similarly $OC \cdot OD=OI_3 \cdot OJ_3$ $\Longrightarrow$ $OI_1 \cdot OI_3 \cdot OJ_1 \cdot OJ_3=OA \cdot OB \cdot OC \cdot OD$ and similarly we obtain $OI_2 \cdot OI_4 \cdot OJ_2 \cdot OJ_4=OA \cdot OB \cdot OC \cdot OD.$ Thus from these relations, we conclude that $I_1,I_2,I_3,I_4$ are concyclic $\Longleftrightarrow$ $OI_1 \cdot OI_3=OI_2 \cdot OI_4$ $\Longleftrightarrow$ $OJ_1 \cdot OJ_3=OJ_2 \cdot JI_4$ $\Longleftrightarrow$ $J_1,J_2,J_3,J_4$ are concyclic.

Lemma 1: In $\triangle ABC$ if $I,E$ are the incenter and A-excenter, we have that \[AI\cdot AE = AB\cdot AC\]Proof: Note that $\triangle ABI \sim \triangle AEC$ due to angle chase, which yields \[\frac{AB}{AI}=\frac{AE}{AC}\Longrightarrow AI\cdot AE = AB\cdot AC \]$\square$ Label the incenters as $I_1,I_2,I_3,I_4$ and the excenters $E_1,E_2,E_3,E_4$, thus we clearly have that $(E_1,I_1,P,I_3,E_3)$ and $(E_4,I_4,P,I_2,E_4)$ are collinear. We now reinterpret the concyclic conditions as \[PI_2\cdot PI_4=PI_1\cdot PI_3\Longleftrightarrow \frac{PI_2\cdot PI_4}{PI_1\cdot PI_3}=1 \quad \quad (1)\]\[PE_2\cdot PE_4=PE_1\cdot PE_3 \Longleftrightarrow \frac{PE_2\cdot PE_4}{PE_1\cdot PE_3}=1 \quad \quad (2)\] I now note that due to Lemma 1, \[\frac{PI_2\cdot PI_4}{PI_1\cdot PI_3}\cdot \frac{PE_2\cdot PE_4}{PE_1\cdot PE_3} = \frac{(PI_2\cdot PE_2)(PI_4\cdot PE_4)}{(PI_1\cdot PE_1)(PI_3\cdot PE_3)} = \frac{(PB\cdot PC)\cdot(PA\cdot PD)}{(PA\cdot PB)(PC\cdot PD)}=1\]Thus, we clearly have that the two concyclic conditions occur if and only if the other one occurs.

Lemma: In a tringle $ABC$ with incenter $I$ and A-excenter $J_A$, we have $AI\cdot AJ_A=AB\cdot AC$. Proof: By the incenter-excenter lemma, there exists a circle $\omega$ through the points $I,J_A,B,C$. Let the second intersection of line $AC$ with $\omega$ be $X$; it follows from $\omega$ being centered on the $A$ angle bisector that $AX=AB$. Our claim is then immediate from PoP. Notice that $I_1,J_1,I_3,J_3$ are all collinear with $O$ since they lie on the line bisecting angles $AOB$ and $COD$. Similarly, $I_2,J_2,I_4,J_4$ are all collinear with $P$ on the line bisecting angles $BOC$ and $DOA$. The given condition is then equivalent to that $$OJ_1\cdot OJ_3=OJ_2\cdot OJ_4\stackrel{?}{\iff} OI_1\cdot OI_3=OI_2\cdot OI_4$$Notice that $OJ_1\cdot OJ_3\cdot OI_1\cdot OI_3=(OJ_1\cdot OI_1)\cdot (OJ_3\cdot OI_3)=OA\cdot OB\cdot OC\cdot OD$ by our lemma, and similarly $OJ_2\cdot OJ_4\cdot OI_2\cdot OI_4=(OJ_2\cdot OI_2)\cdot (OJ4_\cdot OI_4)=OB\cdot OC\cdot OD\cdot OA$. Our desired result immediately follows from the above two products being equal.

Uh the notation is a little different but anyway, We begin with the following lemma Lemma In a triangle $ABC$ with incentre $I$ and excentre $I_A$ \[AI \cdot AI_A = AB \cdot AC \] Proof We use angle chasing and prove that $\Delta AIB \sim \Delta AI_AC$. The claimed lemma follows Label the incenters of $AB,BC,CD,DA$ as $I_1,I_2,I_3,I_4$ and the excentres as $E_1,E_2,E_3,E_4$ Note that \begin{align*} PA\cdot PB\cdot PC \cdot PD &= PA\cdot PB\cdot PC \cdot PD\\ (PA\cdot PB) \cdot (PC \cdot PD) &= (PA\cdot PD) \cdot (PB\cdot PC)\\ (PI_1\cdot PE_1) \cdot (PI_3 \cdot PE_3) &= (PI_4\cdot PE_4) \cdot (PI_2\cdot PE_2)\\ \end{align*} Note that \[PI_1\cdot PI_3 = PI_2 \cdot PI_4 \iff PE_1\cdot PE_3 = PE_2 \cdot PE_4\] However the LHS is true iff the 4 incentres lie on a circle and hence by power of point $P$ we are done!

Version of the problem I received: Rewritten Problem wrote: Diagonals $\overline{AC}$ and $\overline{BD}$ of convex quadrilateral $ABCD$ meet at $P$. Prove that the incenters of the triangles $\triangle PAB, \triangle PBC, \triangle PCD, \triangle PDA$ are concyclic if and only if their $P$-excenters are also concyclic. The key realization is the following: Lemma: In a triangle $\triangle ABC$ with incenter $I$ and $A$-excenter $I_A$, we have $\triangle ABI \sim \triangle AI_AC$. Proof: We clearly have $\angle BAI=\angle I_AAC$. Further, $$\angle AIB=90^\circ+\frac{\angle ACB}{2}=90^\circ+\angle ACI=\angle ACI_A$$since $\angle ICI_A=90^\circ$. $\blacksquare$ From this, we obtain $AB\cdot AC=AI \cdot AI_A$. Now let $I_1,J_1$ denote the incenter and $P$-excenter respectively of $\triangle PAB$, and define $I_2,J_2,I_3,J_3,I_4,J_4$ cyclically. Then by Power of a Point we have \begin{align*} (PI_1)(PI_3)&=(PI_2)(PI_4) & &\iff\\ (PI_1)(PI_3)(PA)(PB)(PC)(PD)&=(PI_2)(PI_4)(PA)(PB)(PC)(PD) & &\iff\\ (PJ_1)(PJ_3)&=(PJ_2)(PJ_4), \end{align*}so $I_1I_2I_3I_4$ is cyclic iff $J_1J_2J_3J_4$ is cyclic. $\blacksquare$

Uhh, similarities have ever been my weakness and it clearly shows how bad I am actually at geometry with this problem, from the time I am dying on this problem for so long . Also, figure is more like Ninja Hattori . We relabel the points for easier naming as in the figure, and label the incenters and excenters too. Now, $\angle PI_1A_1=90^{\circ} +\dfrac{\angle PA_2A_1}{2}=90^{\circ}+\angle PA_2I_1=\angle PA_2J_1$, and $\angle A_1PI_1=\dfrac{A_1PA_2}{2}=\angle J_1PA_2\implies \triangle PA_1I_1\sim\triangle PJ_1A_2\implies\dfrac{PA_1}{PJ_1}=\dfrac{PI_1}{PA_2}\implies PI_1\cdot PJ_1=PA_1\cdot PA_2$. Similarly, we also derive the followings, \begin{align} PI_1\cdot PJ_1=PA_1\cdot PA_2\\ PI_2\cdot PJ_2=PA_2\cdot PA_3\\ PI_3\cdot PJ_3=PA_3\cdot PA_4\\ PI_4\cdot PJ_4=PA_4\cdot PA_1 \end{align} $(1)\times (3)$ and $(2)\times (4)$ gives $PI_1\cdot PJ_1\cdot PI_3\cdot PJ_3=PA_1\cdot PA_2\cdot PA_3\cdot PA_4=PI_2\cdot PJ_2\cdot PI_4\cdot PJ_4$. It is now clear that $I_1I_2I_3I_4$ is cyclic $\iff$ $J_1J_2J_3J_4$ is cyclic.

Let $J_1, J_2, J_3, J_4$ be the four excenters. In general, I claim that in an arbitrary triangle $ABC$, the triangles $ABI$ and $AI_AC$ are similar. This can simply be done by computing $\angle AI_AC = \frac B2$. Thus $$PI_1 \cdot PJ_1 \cdot PI_3 \cdot PJ_3 = PA \cdot PB\cdot PC \cdot PD = PI_2 \cdot PJ_2 \cdot PI_4 \cdot PJ_4,$$which suffices.

Since $\triangle PAI_1\sim\triangle PJ_1B$, we have $PI_1\cdot PJ_1=PA\cdot PB$. Combining this with similar relations for the other sides, we have \[PI_1\cdot PJ_1\cdot PI_3\cdot PJ_3=PA\cdot PB\cdot PC\cdot PD=PI_2\cdot PJ_2\cdot PI_4\cdot PJ_4\]which yields the desired conclusion by PoP. $\square$

Let the incenter of $\triangle PAB$ be $W$ and the excenter be $W_P$. Define points $X$, $Y$, $Z$ similarly. Notice that $\triangle W_PAP \sim \triangle BWP$. By Fact $5$, we have $AWBW_P$ concyclic, and by inscribed angles we have $\angle{AW_PW} = \angle \frac{B}{2} = \angle WBP$. We also have $\angle APW = \angle WPB = \angle \frac{P}{2}$, so we are done by $AA$. From this, we find that $AP \cdot BP = PW \cdot PW_P$. By similar triangles, we have \[\frac{WP}{AP} = \frac{BP}{W_PW} \implies AP \cdot BP = PW \cdot PW_P.\]so it is true. By Power of a Point, we have $PW \cdot PY = PX \cdot PZ \iff PW_P \cdot PY_P = PX_P \cdot PZ_P$ being equivalent to the $P-$excenters being concyclic if and only if the incenters are concyclic. $\newline$ Due to symmetry, we can apply our previous claims to get \[PW \cdot PW_P \cdot PY \cdot PY_P = PX \cdot PX_P \cdot PZ \cdot PZ_P = AP \cdot BP \cdot CP \cdot DP\]\[\implies PW \cdot PY = PX \cdot PZ \iff PW_P \cdot PY_P = PX_P \cdot PZ_P\]So we are done. $\blacksquare$

Define $I_{\triangle}$ and $J_{\triangle}$ to be the incenter and $P$-excenter of $\triangle$, respectively. Claim. $\triangle PAI_{\triangle PAD} \sim \triangle PJ_{\triangle PAD}D$ Proof. Note that $AI_{\triangle PAD}DJ_{\triangle PAD}$ is cyclic as $\angle I_{\triangle PAD}AJ_{\triangle} = \angle J_{\triangle PAD}DI_{\triangle} = 90^{\circ}$. Therefore, \begin{align*} \angle PAI_{\triangle PAD} = I_{\triangle PAD}AD = \angle I_{\triangle PAD}J_{\triangle PAD}D. \end{align*}As $\angle APJ_{\triangle PAD} = \angle J_{\triangle PAD}PD$, then $\triangle PAI_{\triangle PAD} \sim \triangle PJ_{\triangle PAD}D$, as desired. $\blacksquare$ Thus, by similar triangles, then: \begin{align*} \dfrac{AP}{PI_{\triangle PAD}} = \dfrac{J_{\triangle PAD}P}{PD} \quad \longrightarrow \quad AP \cdot PD = J_{\triangle PAD}P \cdot PI_{\triangle PAD}. \end{align*}Thus, \[ \begin{cases} AP \cdot PB &= J_{\triangle PBA}P \cdot PI_{\triangle PBA} \\ BP \cdot PC &= J_{\triangle PCB}P \cdot PI_{\triangle PCB} \\ CP \cdot PD &= J_{\triangle PDC}P \cdot PI_{\triangle PDC} \\ DP \cdot PA &= J_{\triangle PAD}P \cdot PI_{\triangle PAD}. \end{cases} \]Therefore: \begin{align*} AP \cdot BP \cdot CP \cdot PD &= AP \cdot BP \cdot CP \cdot PD \\ J_{\triangle PBA}P \cdot PI_{\triangle PBA} \cdot J_{\triangle PDC}P \cdot PI_{\triangle PDC} &= J_{\triangle PCB}P \cdot PI_{\triangle PCB} \cdot J_{\triangle PAD}P \cdot PI_{\triangle PAD} \\ \left(J_{\triangle PBA}P \cdot PJ_{\triangle PDC}\right) \left(I_{\triangle PBA}P \cdot PI_{\triangle PDC}\right) &= \left(J_{\triangle PCB}P \cdot PJ_{\triangle PAD}\right) \left(I_{\triangle PCB}P \cdot PI_{\triangle PAD}\right) \end{align*}Now, if the incenters are concyclic, then by power of point, $I_{\triangle PBA}P \cdot PI_{\triangle PDC} = I_{\triangle PCB}P \cdot PI_{\triangle PAD}$. So \[ J_{\triangle PBA}P \cdot PJ_{\triangle PDC} = J_{\triangle PCB}P \cdot PJ_{\triangle PAD}, \]which by power of point implies the excenters are concyclic. Instead, if the excenters are concyclic then $J_{\triangle PBA}P \cdot PJ_{\triangle PDC} = J_{\triangle PCB}P \cdot PJ_{\triangle PAD}$, so \[ I_{\triangle PBA}P \cdot PI_{\triangle PDC} = I_{\triangle PCB}P \cdot PI_{\triangle PAD}, \]which implies the incenters are concyclic by power of point. Thus, the incenters are concyclic if and only if the excenters are concyclic, as desired.

Note the well known property that $OA \cdot OB = OI_1 \cdot OJ_1$. The proof is by similarity of triangles $AOI_1, J_1OB$. Similarly, $OI_3 \cdot OJ_3 = OC \cdot OD$. Therefore, $(OI_1 \cdot OI_3) \cdot (OJ_1 \cdot OJ_3) = OA \cdot OB \cdot OC \cdot OD = (OI_2 \cdot OI_4) \cdot (OJ_2 \cdot OJ_4)$. Now $J_1, I_1, O, I_3, J_3$ are collinear, as are $J_2, I_2, O, I_4, J_4$, so $I_1I_2I_3I_4 \text{ cyclic } \iff OI_1 \cdot OI_3 = OI_2 \cdot OI_4 \stackrel{\rm \text{above relation}}{\rm \iff} OJ_1 \cdot OJ_3 = OJ_2 \cdot OJ_4 \iff J_1J_2J_3J_4 \text{ cyclic }$. $\square$