Taking away the restrictions imposed in the problem, i.e. considering 4 points A,B,C,D in general position with no three of them collinear, then the locus will be the union of a conic through A,B,C,D and the lines AC,BD. Using barycentric coordinates WRT $\triangle ABC,$ and denoting $(x:y:z),$ $(p:q:r)$ the coordinates of $P,D,$ the areas of $\triangle PAB,$ $\triangle PBC,$ $\triangle PCD,$ $\triangle PDA$ are given by $\frac{[PAB]}{[ABC]}=\left |\frac{z}{x+y+z} \right| \ , \ \frac{[PBC]}{[ABC]}=\left |\frac{x}{x+y+z} \right|$ $\frac{[PCD]}{[ABC]}=\left |\frac{qx-py}{(x+y+z)(p+q+r)} \right| \ , \ \frac{[PDA]}{[ABC]}=\left |\frac{ry-qz}{(x+y+z)(p+q+r)} \right|$ $[PAB] \cdot [PCD]=[PBC] \cdot [PDA] \Longrightarrow z|qx-py|=x|ry-qz| \Longrightarrow$ $z^2(qx-py)^2=x^2(ry-qz)^2 \Longrightarrow y(rx-pz)(pyz-2qzx+rxy)=0.$ Either $y=0$ $\Longrightarrow$ $P \in AC,$ $rx-pz=0$ $\Longrightarrow$ $P \in BD,$ or $pyz-2qzx+rxy=0$ $\Longrightarrow$ $P$ is on a circumconic $\mathcal{C}$ of $\triangle ABC$ through $D.$ $\mathcal{C}$ is the isogonal of the line $D^*U$ WRT $\triangle ABC,$ where $D^*$ is the isogonal conjugate of $D$ WRT $\triangle ABC$ and $U$ is on $AC,$ such that $U \equiv \left (-\tfrac{r}{c^2}:0:\tfrac{p}{a^2} \right).$

Let $ \angle APB=x_1 $, $ \angle BPC=y_1$, $ \angle CPD=x_2 $, $ \angle DPA=y_2 .$ So, math=inline]$ APB $[/mathmath=inline]$ BPC $[/math$=$math=inline]$ CPD $[/mathmath=inline]$ DPA $[/math $ \Rightarrow $ $ \sin x_1 \sin x_2 = \sin y_1 \sin y_2 $ $ \Rightarrow $ $ \cos(x_1-x_2)=\cos(y_1-y_2) .$ Hence, locus of the point $ P $ is dioganals of the quadriteral $ ABCD $.

I do not understand....