Solution:

a.

If $P=I$ is the incentre of triangle $ABC$, and $r$ its inradius, then it is clear that $A_{1}I = B_{1}I = C_{1}I = 2r$. It follows that $I$ is the circumcentre of $A_{1}B_{1}C_{1}$. On the other hand if $P=I_{1}$ is the excentre of $ABC$ opposite $A$ and $r_{1}$ the corresponding exradius, then again we see that $A_{1}I_{1} = B_{1}I_{1} = C_{1}I_{1} = 2r_{1}$. Thus $I_{1}$ is the circumcentre of $A_{1}B_{1}C_{1}$.

b.
Let $P=O$ be the circumcentre of $ABC$. By definition, it follows that $OA_{1}$ bisects and is bisected by $BC$ and so on. Let $D, E, F$ be the mid-points of $BC, CA, AB$ respectively. Then $FE$ is parallel to $BC$. But $E, F$ are also mid-points of $OB_{1}, OC_{1}$ and hence $FE$ is parallel to $B_{1}C_{1}$ as well. We conclude that $BC$ is parallel to $B_{1}C_{1}$. Since $OA_{1}$ is perpendicular to $BC$, it follows that $OA_{1}$ is perpendicular to $B_{1}C_{1}$. Similarly $OB_{1}$ is perpendicular to $C_{1}A_{1}$ and $OC_{1}$ is perpendicular to $A_{1}B_{1}$. These imply that $O$ is the orthocentre of $A_{1}B_{1}C_{1}$. (This applies whether $O$ is inside or outside $ABC$.)

c.
Let $P=H$, the orthocentre of $ABC$. We consider two possibilities; $H$ falls inside $ABC$ and $H$ falls outside $ABC$.

Suppose $H$ is inside $ABC$; this happens if $ABC$ is an acute triangle. It is known that $A_{1}, B_{1}, C_{1}$ lie on the circumcircle of $ABC$. Thus $\angle C_{1}A_{1}A = \angle C_{1}CA = 90^{\circ} - A$. Similarly $\angle B_{1}A_{1}A = \angle B_{1}BA = 90^{\circ} - A$. These show that $\angle C_{1}A_{1}A = \angle B_{1}A_{1}A$. Thus $A_{1}A$ is an internal bisector of $\angle C_{1}A_{1}B_{1}$. Similarly we can show that $B_{1}B$ bisects $\angle A_{1}B_{1}C_{1}$ and $C_{1}C$ bisects $\angle B_{1}C_{1}A_{1}$. Since $A_{1}A, B_{1}B, C_{1}C$ concur at $H$, we conclude that $H$ is the incentre of $A_{1}B_{1}C_{1}$.

OR If $D, E, F$ are the feet of perpendiculars of $A, B, C$ to the sides $BC, CA, AB$ respectively, then we see that $EF, FD, DE$ are respectively parallel to $B_{1}C_{1}, C_{1}A_{1}, A_{1}B_{1}$. This implies that $\angle C_{1}A_{1}H = \angle FDH = \angle ABE = 90^{\circ} - A$, as $BDHF$ is a cyclic quadrilateral. Similarly, we can show that $\angle B_{1}A_{1}H = 90^{\circ} - A$. It follows that $A_{1}H$ is the internal bisector of $\angle C_{1}A_{1}B_{1}$. We can proceed as in the earlier case.

If $H$ is outside $ABC$, the same proofs go through again, except that two of $A_{1}H, B_{1}H, C_{1}H$ are external angle bisectors and one of these is an internal angle bisector. Thus $H$ becomes an excentre of triangle $A_{1}B_{1}C_{1}$.