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 $u+x+y+z.$