We first prove the “if” direction by assuming that $P$ lies on $\Omega$ and triangles $PBD$ and $PCE$ have the same incenter $I$. We consider the diagram shown above. (We can adjust our proof easily for other possible configurations.) Because $PI$ bisects both $\angle BPD$ and $\angle EPC$, $\angle EPB = \angle DPC$. Because $ABCP$ is cyclic, $\angle PBE = \angle PBA = \angle PCA = \angle PCD$. Thus, we have $\angle EPB = \angle DPC$ and $\angle PBE = \angle PCD$, so triangles $PEB$ and $PDC$ are similar. In other words, there is a spiral similarity centered at $P$ sending triangle $PDC$ to triangle $PEB$. Therefore, there is a spiral similarity centered at $P$ sending triangle $PED$ to triangle $PBC$. In particular, we have
$$\angle DEP = \angle CBP \quad \text{and} \quad \angle PDE = \angle DCB = \angle ACB,$$
from which it follows that
$$\begin{align*} \angle EPC &= \angle EPD - \angle DPC = 180^\circ - \angle DEP - \angle PDE + 180^\circ - \angle CDP - \angle PCD \\ &= (360^\circ - \angle PDE - \angle CDP) - \angle DEP - \angle PCD \\ &= \angle EDC - \angle CBP - \angle PCA. \end{align*}$$
Because $BCDE$ and $ABCP$ are cyclic, we obtain
$$\begin{align*} \angle EPC &= \angle EDC - \angle CBP - \angle PCA \\ &= 180^\circ - \angle CBE - \angle CBP - \angle PBA \\ &= 180^\circ - \angle CBE - \angle CEP = 180^\circ - 2\angle CBE. \end{align*}$$
Because $O$ is the circumcenter of acute triangle $BEC$, $\angle COE = 2\angle CBE$. Thus, $\angle EPC = 180^\circ - 2\angle CBE = 180^\circ - \angle COE$ or $\angle EPC + \angle COE = 180^\circ$, implying the $CPEO$ is cyclic.
In exactly the same way, we can show that $BPDO$ is cyclic. Now consider the three cyclic quadrilaterals $BCDE$, $BPDO$, $CPEO$. Their pairwise radical axes are lines $BD$, $PO$, $CE$, from which it follows that lines $BD$, $PO$, $CE$ meet at the radical center of the three circumcircles.

First, we prove the “only if” direction. We define $P_1$ and $I$ as in the second part of the first solution. The discussion of the configuration remains the same. It remains only to present the proofs of the two steps.

Step 1: Because $OX \cdot OP_1 = OE^2 = OC^2$, triangles $OEX$ and $OP_1E$ are similar and triangles $OCX$ and $OP_1C$ are similar. Thus, by also noting that triangle $EOC$ is isosceles, we see that
$$\angle EP_1O = \angle OEX = \angle OEC = \angle ECO = \angle XCO = \angle OP_1C.$$
Therefore, $OEP_1C$ is cyclic, and $OP_1$ bisects $\angle EPC$. Hence, we have $OI = OC = OE$. It is well known that $I$ is thus the incenter of triangle $P_1CE$. Here, we use the fact that if triangle $MNK$ is inscribed in a circle $\delta$, then its incenter lies on the circle centered at the midpoint $Z$ of the arc $NK$ not containing $M$ with radius $ZN = ZK$. In the exact same way, we can show that $I$ is the incenter of triangle $P_1BD$. We conclude that triangles $P_1BD$ and $P_1CE$ share the same incenter.

Step 2: Because $BP_1DO$ and $CP_1EO$ are cyclic, we have $\angle BP_1O = \angle BDO$ and $OP_1C = \angle OEC$, and so
$$\angle BP_1C = \angle BP_1O + \angle OP_1C = \angle BDO + \angle OEC.$$
On the other hand, because $O$ is the circumcenter of triangles $BDC$ and $BEC$, we have $\angle BDO = \angle OBD = 90^\circ - \angle DCB = 90^\circ - \angle ACB$ and $\angle OEC = \angle ECO = 90^\circ - \angle CBE = 90^\circ - \angle CBA$. Combining the last three identities gives
$$\angle BP_1C = \angle BDO + \angle OEC = 90^\circ - \angle ACB + 90^\circ - \angle CBA = \angle BAC,$$
implying that $A$, $B$, $C$, and $P_1$ lie on a circle, and $P = P_1$. This completes the proof of the “only if” direction.

We now prove the “if” direction by assuming that $P$ satisfies the following property $\mathcal{P}$:
$P$ lies on $\Omega$ and triangle $PBD$ and $PCE$ have the same incenter.

As shown in the first part of the first solution, $\angle EPD = \angle BAC = \angle EAD$, implying that $AEDP$ is cyclic. That is, $P$ is the intersection of the circumcircles of triangles $ABC$ and $ADE$ other than $A$. This intersection is unique. On the other hand, by the "only if" direction, the intersection of ray $OX$ and $\Omega$ also satisfies property $\mathcal{P}$. By the uniqueness of the two constructions, we conclude that $P$ is the intersection of ray $OX$ and $\Omega$, completing our proof.