Let $A_3, B_3, C_3$ be the feet of the perpendiculars from $P$ to the sides $BC, CA, AB$ respectively. Clearly, $A', B', C'$ are on the lines $PA_3, PB_3, PC_3$, respectively. By the Pythagoras Theorem, we have
$$A'P^2 - A'A_3^2 = A'A_1^2 - A'A_3^2 = A_1P^2 - A_3P^2 = A_1P^2 - (A_3A' + A'P)^2$$
and hence $PA' \cdot PA_3 = PA_1^2/2$. Similarly we get that $PB' \cdot PB_3 = PB_1^2/2$ and $PC' \cdot PC_3 = PC_1^2/2$. Therefore we conclude that $PA' \cdot PA_3 = PB' \cdot PB_3 = PC' \cdot PC_3$. Hence the quadrilaterals $A_3C_3C'A', B_3A_3A'B', C_3B_3B'C'$ are cyclic. By the sine law in triangles $A_3B_3A'$ and $AC_3A'$, we have

$$\frac{A'B_3}{AA'} = \frac{\sin \angle A'AB_3}{\cos \angle A'B_3P} \quad \text{and} \quad \frac{A'C_3}{AA'} = \frac{\sin \angle A'AC_3}{\cos \angle A'C_3P}$$
Hence we get
$$\frac{\sin \angle A'AB_3}{\sin \angle A'AC_3} = \frac{A'B_3}{A'C_3} \cdot \frac{\cos \angle A'B_3P}{\cos \angle A'C_3P}$$
Similarly, we get that
$$\frac{\sin \angle B'BC_3}{\sin \angle B'BA_3} = \frac{B'C_3}{B'A_3} \cdot \frac{\cos \angle B'C_3P}{\cos \angle B'A_3P}$$
and
$$\frac{\sin \angle C'CA_3}{\sin \angle C'CB_3} = \frac{C'A_3}{C'B_3} \cdot \frac{\cos \angle C'A_3P}{\cos \angle C'B_3P}$$
By the converse of the Trigonometric Ceva Theorem, in order to show that $AA', BB'$ and $CC'$ are concurrent it is sufficient to show that
$$\frac{A'B_3}{A'C_3} \cdot \frac{B'C_3}{B'A_3} \cdot \frac{C'A_3}{C'B_3} \cdot \frac{\cos \angle A'B_3P}{\cos \angle A'C_3P} \cdot \frac{\cos \angle B'C_3P}{\cos \angle B'A_3P} \cdot \frac{\cos \angle C'A_3P}{\cos \angle C'B_3P} = 1$$
By the concyclicity, $\angle A'C_3P = C'A_3P$, $B'A_3P = A'B_3P$, $C'B_3P = B'C_3P$ and hence
$$\frac{\cos \angle A'B_3P}{\cos \angle A'C_3P} \cdot \frac{\cos \angle B'C_3P}{\cos \angle B'A_3P} \cdot \frac{\cos \angle C'A_3P}{\cos \angle C'B_3P} = 1$$
Now, we have to show that
$$\frac{A'B_3}{A'C_3} \cdot \frac{B'C_3}{B'A_3} \cdot \frac{C'A_3}{C'B_3} = 1$$
By concyclicity again, the triangles $PA_3C'$ and $PC_3A'$ are similar and hence
$$\frac{C'A_3}{A'C_3} = \frac{PA_3}{PC_3}$$
and similarly we get that
$$\frac{A'B_3}{B'A_3} = \frac{PB_3}{PA_3} \quad \text{and} \quad \frac{B'C_3}{C'B_3} = \frac{PC_3}{PB_3}$$
Side by side product of last three equalities yields
$$\frac{A'B_3}{A'C_3} \cdot \frac{B'C_3}{B'A_3} \cdot \frac{C'A_3}{C'B_3} = \frac{PA_3}{PC_3} \cdot \frac{PB_3}{PA_3} \cdot \frac{PC_3}{PB_3} = 1$$
as desired.