As an alternative to mass points or affine transformations,
you could also use an almost purely algebraic approach. Let
$a$ be the area of the top gray triangle, let $b$ be the area
of the middle gray triangle, and $c$ the area of the bottom gray
triangle.
The fact that $|EA|/|EC| = K(BEA)/K(BCE) = K(PEA)/K(PEC)$
is expressed algebraically as
$$
\frac{a+b+x}{c+y+z} = \frac{x}{y}.
$$
The fact that $|BD|/|DC| = K(BDA)/K(DCA) = K(BDP)/K(DCP)$
is expressed algebraically as
$$
\frac{a+b+c}{x+y+z} = \frac{c}{z}.
$$
Defining the additional variables $u=a+b+c$ and $v=a+b$ (making $c=u-v$)
makes this a system of two linear equations in the two unknowns $u$ and $v$:
$$
\begin{align}
(v+x)y &= x(u-v+y+z) \\
uz &= (u-v)(x+y+z).
\end{align}
$$
Solving for $u$ yields
$$
u = \frac{xz(x+y+z)}{xy+y^2-xz}.
$$
The total area is $u+x+y+z.$