Solution:
Let $K$ and $M$ be the tangency points of the circle with $AP$ and $CP$ respectively, $L$ the point where it touches the circumscribed circle of $ABC$, $T$ the middle of the arc $AB$ of the circumscribed circle, $I$ the center of the inscribed circle of $ABC$.

The tangent to the circle $ABC$ at $T$ is parallel to the line $AB$. So the similarity transformation ("stretching") centered at $K$ taking the circle $ABC$ to the circle $KLM$ takes $AB$ to this tangent, and hence $T$ to $L$. So $K$, $L$ and $T$ are collinear. We now prove that points $K$, $M$ and $I$ are collinear. Let $M'$ be the point of intersection of the line $KI$ with the circle $KLM$. We want to show $M = M'$.

First, let's note that the quadrilateral $LCIM'$ can be inscribed into a circle. In fact, $\angle LCT$ is equal to the half-sum of the $\operatorname{arcs} LA$ and $AT$, which is equal to $\frac{1}{2}(\breve{LA} + TB) = \angle LKA$. But $LKA$ is the angle formed by the tangent $AK$ and the chord $LK$ of the circle $LKM$, and is therefore equal to $LM'K$, subtended by this chord. Hence $\angle LCT = \angle LM'K$ and the quadrilateral $LCMI$ can be inscribed in a circle. Therefore $\angle LM'C = \angle LIC$.

Further, $\angle AKT = \frac{1}{2}(\breve{AT} + \breve{BL}) = \frac{1}{2}(\check{TB} + \check{BL}) = \angle LAT$. Hence the triangles $TAK$ and $TLA$ are similar, and so $TK \cdot TL = TA^2$. Also $\angle AIT = \angle CAI + \angle ACI = \angle CAI + \angle TAB = \angle BAI + \angle TAB = \angle TAI$. Therefore $TAI$ is an equilateral triangle, and so $TI^2 = TA^2 = TK \cdot TL$, which implies that the triangles $TKI$ and $TIL$ are similar, and so $\angle LIC = \angle LKI$. Hence $\angle LM'C = \angle LIC = \angle LKM'$, so the angle subtended by the chord $LM'$ is equal to the angle between this chord and $M'C$, so $M'C$ is a tangent, and so $M' = M$.

It is time to use the definition of the point $P$. Draw a tangent to the inscribed circle of $ABC$ parallel to $AB$. Let it be tangent to this circle at point $F$. The circle tangent to the continuations of $AC$, $BC$ and to $AB$ at $P$ can be obtained from the inscribed circle of $ABC$ by a similarity transformation centered at $C$. This transformation would take $F$ to $P$. Hence $C$, $F$ and $P$ are collinear. Let $V$ be the midpoint of $PE$, where $E$ is the tangency point of $AB$ with the inscribed circle of $ABC$. As $IV \parallel FP$ ($IV$ connects the midpoints of the sides of $FEP$), $KIV$ is similar to $KMP$, and so $VI = VK$. Hence $IE^2 = VI^2 - VE^2 = VK^2 - VE^2 = (VK + VE)(VK - VE) = EK \cdot PK$.

Now let $D$ be the center of the circle $KLM$. As $\angle KDP = \angle IKE$, the triangles $KDP$ and $EKI$ are similar. Hence $DK \cdot IE = EK \cdot PK$, and so $IE^2 = EK \cdot PK = DK \cdot IE$, and so $IE = DK$, as wanted.