We claim that the answer is a circle concentric with $\Omega$, with radius $\frac{2R-r}{2}$, if $R, r$ are the radii of $\Omega, \omega$. Let $R, O$ be the centers of $\Omega, \omega$, and let $C = IS \cap (AIB)$; then the orthocenter of $\triangle{ABC}$ is the reflection of $I$ over $S$, so by the Euler Line, if $M$ is the midpoint of $AB$ then $MO = \frac{SC}{2} - SI$. By Power of a Point, $SA \cdot SB = SI \cdot SC$, so $SC = \frac{SA \cdot SB}{SI} \implies MO = \frac{SA \cdot SB - SI^2}{2 \cdot SI}$. $SA \cdot SB = RP^2 - RS^2$, so $OR = MO - MR = \frac{RP^2 - RS^2 - PI^2}{2 \cdot PI} - MR = \frac{RP^2 - PI^2}{2 \cdot PI} - \frac{RS^2 + 2 \cdot RM \cdot PI}{2 \cdot PI}$. By the Pythagorean Theorem, $RS^2 = RI^2 + RM^2 - (RM + SI)^2 \implies RS^2 + 2 \cdot RM \cdot SI = RI^2 - SI^2$, so $OR = \frac{RP^2 - RI^2}{2 \cdot PI}$ which is a fixed value, simplifying to $\frac{RP + RI}{2}$ by difference of squares. $\square$

See post #4

You can solve it by law of sine

Let $O,O'$ be centers of $PAB$ and $IAB$ and let $R,r$ be radius of $PAB$ and $IAB$. Note that $P_{PAB}(I)-P_{IAB}(I) = 2.IS.OO'$ and both $P_{PAB}(I)-P_{IAB}(I)$ and $IS$ are fixed so $OO'$ is also fixed so $O'$ lies on a circle centered at $O$. Now note that $P_{PAB}(I)-P_{IAB}(I) = r(2R-r) = 2.r.OO'$ so the radius of this circle is $\frac{2R-r}{2}$.