Note that $P$ is the orthocenter of $\triangle AP_bP_c$, let $Q$, $R$ be the reflection of $P$ over $AC$ and $AB$ respectively. Introduce the orthocenter $H$ and circumcenter $O$ of $\triangle ABC$ and let $T = \overrightarrow{AO} \cap (AP_bP_c)$. \[\measuredangle P_bQT = \measuredangle P_bAT = 90^\circ - \measuredangle AA'B = 90^\circ - \measuredangle ACP = \measuredangle CPQ = \measuredangle P_bQC\]So $Q,C,T$ are collinear and similarly $R,B,T$ are collinear. Again, $\measuredangle CTA = \measuredangle QP_bA = \measuredangle HBA = \measuredangle ACH = \measuredangle AP_cR = \measuredangle ATB$ so $\measuredangle CTB = \measuredangle COB$ meaning $T=\overrightarrow{AO} \cap (BOC)$ which is fixed.

Let $A'$ be the antipode of $A$ on $(ABC)$ and $BA'\cap AC=F, CA'\cap AB=E, AA'\cap EF=D$. Let $K$ be the altitude from $A$ to $BC$. We will use the method of moving points. Animate $P$ on $BC$. Let $\phi$ be the inversion centered at $A$ with radius $AD$ and $,{BB_1}_{\infty},{CC_1}_{\infty}$ be the points in infinity which are on altitudes from $B,C$. \[f: P\mapsto P(CC_1)_{\infty}\cap AC=P_C\mapsto \phi(P_C)\mapsto \phi(P_C)D\cap AB\]\[g:P\mapsto P(BB_1)_{\infty}\cap AB=P_B\mapsto \phi(P_B)\]These polynomials have degree $2$. $\bullet $ For $P=B$, $\ f:P\mapsto F\mapsto \phi(F)D\cap AB$ and $g:P\mapsto B\mapsto \phi(B)$ They are same since $A,B,D,F$ are cyclic $\iff \phi(B),D,\phi(F)$ are collinear. $\bullet $ For $P=C$, $\ f:P\mapsto C\mapsto \phi(C)D\cap AB$ and $g:P\mapsto \phi(E)$ They are same since $A,C,D,E$ are cyclic $\iff \phi(C),D,\phi(E)$ are collinear. $\bullet $ For $P=K$, $\ K\mapsto X\mapsto \phi(X)D\cap AB$ and $g:K\mapsto Y\mapsto \phi(Y)$ Since $ABKC\sim AFDE$, we have $\frac{CY}{YF}=\frac{CK}{KB}=\frac{ED}{DF}$ Thus, $X,Y$ are the perpendiculars from $D$ to $AB,AC$ respectively. This gives that $ADXY$ is cyclic hence $\phi(Y),D,\phi(X)$ are collinear. These $3$ cases give the desired result.$\blacksquare$