Problem:
$ ABCD $ is a tangential quadrilateral and $ P $ is a point such that $ PA=PC, PB=PD . $ Let $ I_1, I_2, I_3, I_4 $ be the incenter of $ \triangle PDA, \triangle PAB, \triangle PBC, \triangle PCD $, respectively $ . $
Prove that $ I_1, I_2, I_3, I_4 $ are concyclic
An interesting and similar problem can be found here, with solution: http://forumgeom.fau.edu/FG2011volume11/FG201108.pdf.
Proof of the Conjecture from the Hyacinthos Group $9022$:
Reproducing here two associated theorems and their proofs to prove the Christopher Bradley Conjecture:-
$\varsigma 1$. $\color{green}{\text{Triangle Lemmata:}}$
Consider $\triangle ABC$ and a point $P$ on side $AC$. Let the incircles of $\triangle PAB$ and $\triangle PBC$ have centres $X$ and $Y$ respectively. The incircle of $\triangle ABC$ touches $AC$ at $U$. Then,
Theorem $1$: The points $P$ and $U$ lie on the circle with diameter $XY$.
Theorem $2$: The point $U$ lies on the internal common tangent of the incircles of triangles $PAB$ and $PBC$ different from the line $BP$.
$\color{red}{\text{Proof of theorem 2:}}$ Let $U$ be the point where the internal common tangent of the incircles entered at $X$ and $Y$ different from $BP$ meets $AC$. We will show that the incircle of $\triangle ABC$ touches $AC$ at $U$.
Let the internal common tangents of the incircles of triangles $PAB$ and $PBC$ touch these circles at $X'$ and $Y'$ respectively. As the internal common tangents to two circles have equal length, we get $X'Y'=X''Y''$ where $X''$ and $Y''$ are points where the incircles of triangles $PAB$ and $PBC$ touch $BP$. Thus, we have, $$PX''-PY''=X''Y''=X'Y'= UY'-UX'$$ If $J$ and $K$ are the points where the incircles of triangles $PAB$ and $PBC$ touch $AC$, then the fact that the two tangent segments from a point to the circle are equal in length gives us $PX''=PJ, PY''=PK,UX'=UJ,UY'=UK$. Using the fact that $PJ=KJ-PK$ and $UK=KJ-UJ$ gives us $PK=UJ$. Thus, $$AU=AJ+UJ=AJ+PK=(AB+AP-BP)/2 + (BP+CP-BC)/2 = (AB + AC-BC)/2$$ and $U$ is the point of tangency to the incircle of $\triangle ABC$ with side $AC$. This proves the theorem.
$\color{red}{\text{Proof of theorem 1:}}$ Now the points $X$ and $Y$ lie on the two angle bisectors of the angle formed by the lines $AC$ and $UX'Y'$ (because both of these lines touch the two incircles). Hence, $\angle{XUY} = 90^\circ$ and $U$ lies on the diameter $XY$. Using the same reasoning, for line $BP$ instead of $UX'Y'$ proves the other part, ultimately proving the theorem.
$\varsigma 2.\color{green}{\text{The Conjecture:}}$
Let $ABCD$ be a circumscribed quadrilateral with the incentre $O$. The diagonals $AC$ and $BD$ meet at $P$. Let $X,Y,Z,W$ be the incentres of the triangles $PAB, PBC, PCD, PDA$ respectively and $(X), (Y), (Z), (W)$ be the respective incircles. Then we have to prove that $X, Y, Z, W$ are concyclic.
$\color{red}{\text{Proof of the Conjecture:}}$ Call $(O)$ the incircle of the quadrilateral $ABCD$, its centre being $O$. Let $(A'), (B'), (C'), (D')$ be the incircles of triangles $DAB, ABC, BCD, CDA$ where $A', B', C', D'$ are their respective centres. We make use of a well-known fact about circumscribed quadrilaterals that: the incircles $(A')$ and $(C')$ of triangles $DAB$ and $BCD$ touch $BD$ at one point $T$.
The incircles $(B')$ and $(D')$ of triangles $ABC$ and $CDA$ touch $AC$ at one point $U$. The point $T$ lies on $A'C'$; moreover the points $T$ and $R$ divide $A'C'$ harmonically, since $T$ is the internal centre of similitude of $(A')$ and $(C')$ and $R$ their external centre of similitude.
Now we can apply theorem $1$ to $\triangle ABC$ and the point $P$ on the side $AC$. Thus, points $P$ and $U$ lie on the circle with diameter $XY$. Hence, the points $X, Y, P, U$ are concyclic and $RX\times RY = RP\times RU$. Similarly, $RZ\times RW = RP\times RU$, and thus, $RX\times RY =RZ\times RW$, and the points $X,Y,Z,W$ are concyclic. This proves the theorem.
Hope it helps.