Up to an affine transformation, without loss of generality (wlog) $P=(0,0),A=(1,0),C=(0,1)$ and $B=(u,v)$ for some real $u,v$. (This is because any nonsingular affine transformation changes all areas proportionally and preserves all affine combinations: $f((1-s)P+sQ)=(1-s)f(P)+sf(Q)$ for any affine transformation $f$, any points $P$ and $Q$, and any real $s$).
Similarly to the previous comment by the OP, we now understand the condition $t=\frac{\overline{PA'}}{\overline{PA}}$ together with the homothety condition as $A'=tA,B'=tB,C'=tC$, so that
$$A'=(t,0),\quad B'=(tu,tv),\quad C'=(0,t).$$
Since $A''$ is collinear with $B$ and $C$, we have
$$A''=aB+(1-a)C=(a u, 1 +a(v-1))$$
for some real $a$. Consider now the vector
$$V:=A''-A'=(a u-t,1 +a(v-1)).$$
The parallelity condition implies that
$$B''=B'+bV=b'A+(1-b')C$$
for some real $b,b'$. Solving the latter vector equation for $b,b'$, we get
$$B''=\Big(\frac{t (-a u+u-1)+a u+t^2 v}{a (u+v-1)-t+1},\frac{-a (t u+v-1)+t^2 v+t u-1}{-a
(u+v-1)+t-1}\Big).$$
Similarly,
$$C''=\Big(\frac{(t-a u) (t (u-1)+v)}{-a (u+v-1)+t v+u-1},\frac{v \left(-a (t
u+v-1)+t^2-1\right)}{-a (u+v-1)+t v+u-1}\Big).$$
The signed area of the triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ in $\Bbb R^2$ is
$$\frac12\det\begin{pmatrix}
1&x_1&y_1\\
1&x_2&y_2\\
1&x_3&y_3
\end{pmatrix}.\tag{10}\label{10}$$
So, the respective signed areas of the triangles $A''B''C''$ and $ABC$ are
$$-\tfrac12\,t (t+1) (u + v-1)\quad\text{and}\quad \tfrac12\,(u + v-1),$$
which completes the proof.
