If $P\ \equiv\ B$, then the point $G$ coincides with the tangency point $D$ of the incircle with $BC$. Incircle, center $I$ and $ID \bot BC\ ,\ IE \bot AC$. If $P\ \equiv\ C$, then $K\ \equiv\ I$ and $G\ \equiv\ S$. Locus is the line segment $DS$ and $S \in DE$.

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Let $I,J$ be incenter of $\triangle ABC,\triangle APC$, respectively ; $E$ be the tangency point of incircle of $\triangle APC$ with segment $AP$ ; $T = EF \cap CJ$. Observe that: 1. $\angle AKP = 90^\circ + \frac{\angle B}{2}$ is fixed. 2. Points $C,I,J$ are collinear on $\angle C$ bisector. 3. Both $PK,EF$ are perpendicular to $PJ$, so $PK \parallel EF$. Hence $G \in EF$. 4. By IRAN LEMMA in $\triangle APC$, we get $T$ lies on $\odot(AC)$ and homothety of line $BC$ through $A$ by a factor of $\frac{1}{2}$. Basically, $T$ is fixed regardless of $P$ (and $T$ has several properties of the IRAN LEMMA wrt $\triangle ABC$). So our problem can be restated as follows (notations are the same as above): Restated problem wrote: Point $K$ moves on $BI$. $G$ is the intersection of $CK$ and the line through $T$ which makes an angle of $90^\circ + \frac{\angle B}{2}$ with line $AK$ (angles are directed). Find the locus of $G$. Claim 1: $G$ either moves on a conic passing through $C$ and $T$ ; or $G$ moves on a line. Proof: Consider any four positions of $G$, say $G_1,G_2,G_3,G_4$. We have that, \begin{align*} C(G_1,G_2 ; G_3,G_4) &= (K_1,K_2 ; K_3,K_4) \qquad \qquad ~(\text{projecting onto line } BI)\\ &= A(K_1,K_2 ; K_3,K_4) \qquad \qquad (\text{projecting through } A) \\ &= T(G_1,G_2 ; G_3,G_4) \qquad \qquad (\text{rotating the lines by angle } 90^\circ + \frac{\angle B}{2} ) \end{align*}Our Claim just follows by the projective definition of a conic. $\square$ Let the incircle of $\triangle ABC$ touch sides $CA,CB$ at points $X,Y$, respectively. We will show that $G$ moves on line $XY$. Consider the following three special cases: (i): $K=B$. Then $G = Y$. (ii): $K = BI \cap AC$. Then $G = X$. (iii): $K = BI \cap XY$. Then $G = BI \cap XY$. Note that all three cases are non-trivial (they are in increasing order of difficulty) and we need IRAN LEMMA (wrt $\triangle ABC$) in all of them. So we get three different positions of $G$ lie on line $XY$. Now since a Conic cannot contain three collinear points, so using our Claim 1 we conclude that $G$ moves on line $XY$. This completes the proof. $\blacksquare$ Remark 1: There is a slight flaw in the above solution. When $BA = BC$, then case (ii), (iii) are same. But we can fix that. The analysis is enough motivation for us to conjecture that lines $XY$ is the desired locus. Now through $C,T$, we are creating two projective maps from line $BI$ to lines $XY$. We want these two projective maps to coincide, so it suffices to show they coincide for three different positions of $K$. So this time, we can consider the third special case as $K=I$. Remark 2: The official solution is a elementary proof.