Consider the problem in complex plane with the circumcircle of the given circle is the unit circle. Let $O_1,O_2$ denote the circumcenter of $\triangle{ABP},\triangle{ACP}$. Let the complex number $a,b,c,d,p,o_1,o_2$ correspond to points $A,B,C,D,P,O_1,O_2$, respectively. We've $p=\frac{ac(b+d)-bd(a+c)}{ac-bd}$. This gives $o_1=\frac{\begin{vmatrix} a & 1 & 1 \\ b & 1 & 1 \\ \frac{ac(b+d)-bd(a+c)}{ac-bd} & \frac{ac(b+d)-bd(a+c)}{ac-bd} \times \frac{a+c-b-d}{ac-bd} & 1 \end{vmatrix}}{\begin{vmatrix} a & \frac{1}{a} & 1 \\ b & \frac{1}{b} & 1 \\ \frac{ac(b+d)-bd(a+c)}{ac-bd} & \frac{a+c-b-d}{ac-bd} & 1 \end{vmatrix}}=\frac{\frac{1}{(ac-bd)^2}\times \Big( (a-b)^2(a-d)(c-b)(c-d)\Big)}{\frac{1}{ac-bd}\times \Big( \frac{(a-b)^2(a-d)(c-b)}{ab}\Big) }=\frac{ab(c-d)}{ac-bd}$. Similarly, $o_2=\frac{cd(a-b)}{ac-bd}$. Note that we've $|a-b|=\ell \implies \ell^2=(a-b)(\frac{1}{a}-\frac{1}{b})\implies \frac{a}{b}+\frac{b}{a}=2-\ell^2$. Let $t$ be a complex number that $t+\frac{1}{t}=2-\ell^2$. Note that the values of $a,b\in \mathbb{C}$ that $\frac{a}{b}=t$ is enough to cover all points $A,B$ on the circle that $|AB|=\ell$. So, we've the condition $\frac{a}{b}=t$ where $t$ is independent to the choices of $a$. Note that $|ac-bd|^2=(ac-bd)(\frac{1}{ac}-\frac{1}{bd})=2-\frac{bd}{ac}-\frac{ac}{bd}=2-\frac{d}{ct}-\frac{tc}{d}$. Also, we've $|a|=|b|=|c|=|d|=1$ and $|a-b|=\ell$ and $|c-d|$ is fixed, i.e. independent to the choices of $a$. So, $|o_1|=\Big| \frac{ab(c-d)}{ac-bd}\Big|$ and $|o_2|=\Big| \frac{cd(a-b)}{ac-bd}\Big|$ are both independent to the choices of $a$. This means $O_1$ lie on a fixed circle, i.e. the circle centered at the origin with radius $\Big| \frac{ab(c-d)}{ac-bd}\Big|$. The similar locus holds for $O_2$. Hence, the answer is both locus are circle concentric with the given circle.

let $O,I,I'$ be the centers of the first circle and the circumcircles of $ABP,BCP$ angle chase leads to $\angle BAP $ is constant then the triangles $AIB$ remain congruent also $OAB$ thus $OI$ is constant hence the locus of I is circle. but the triangles $BCP$ are similar so $\frac{CI'}{CB}\overset{\text {note it }}{=}k$ remains constant and also $\angle BCI'\overset{\text {note it }}{=}\alpha$ remains constant which means that $B$ is sent to $I'$ by the similarity with center $C$ ,ratio $k$ and and angle $ \alpha$ therfore the locus of $I'$ is also circle . RH HAS

Let $T$ be the circumcenter of $\triangle BCP$. Now $\angle TBC = \angle TCB$ is fixed. Let $X = \overline{TB} \cap \text{circle} \ne B$. Now as $C$ and $\angle XBC$ is fixed, so $X$ is fixed. Now $\angle XTC = 2 \angle XBC$ is fixed. Hence $T$ lies on a circle passing through $X,C$. $\blacksquare$