Proof:

We construct some points:
O = circumcenter of ABC
A' = midpoint of AP
Cp = reflection of Pc in midpoint of AB
Bp = reflection of Pb in midpoint of AC
Y is a point such that PPcCpY is a rectangle
Z is a point such that PPbBpZ is a rectangle

Because some reflections:
PY = PcCp = ABa
PZ = PbBp = ACa
Moreover PY//ABa and PZ//ACa therefore triangle ABaCa is reflection of triangle PYZ in A' midpoint of AP or (Oa) = reflection of circumcircle PYZ in A'.

By constructions we can show:
Y = reflections of P in perpendicular bisector of AB
Z = reflections of P in perpendicular bisector of AC
therefore center of circumcircle PYZ is circumcenter O of ABC. The circumcircle PYZ is centered at O and passing P and we denote it as (OP) then:
(Oa) = reflection of (OP) in A' midpoint of AP.

Similarly we can show that:
(Ob) = reflection of (OP) in B' midpoint of BP
(Oc) = reflection of (OP) in C' midpoint of CP

So three circles (Oa), (Ob), (Oc) are congruent with radius OP

Some remarks:
a). This fact is generalized of "three congruent circles by reflection" in altitudes and in angle bisector:
- in altitude case P = orthocenter
- in angle bisector case P = incenter (with regard that if IaIbIc is pedal triangle of incenter I then: reflection in IA of reflection in IC of B is reflection in IIc of B)
b). In its order, be course OAOaP is parallelogram, this fact is also one special case (when one point is circumcenter) of "Infinite Of Congruent Circles From Translations"

Best regards,
Bui Quang Tuan