Consider all occurring points as vectors, as in @Calvin Lin's answer, and write $\mu$ for ${1\over3}$. Then $$p=(1-\mu)a+\mu b,\quad h=(1-\mu)d+\mu c,\quad n=(1-\mu)a+\mu d,\quad e=(1-\mu) b+\mu c\ .$$ It follows that $$(1-\mu)p+\mu h=(1-\mu)n+\mu e\quad(=:w')\ ,$$ which shows that in fact $$w=w'=(1-\mu)^2 a +\mu(1-\mu)(b+d)+\mu^2 c\ .$$ Interchanging $a$ and $c$ here gives $$y=(1-\mu)^2 c +\mu(1-\mu)(b+d)+\mu^2 a\ ,$$ so that we arrive at $$w-y=(1-2\mu)(a-c)\ .$$ Appealing to symmetry again we conclude that we also have $$x-z=(1-2\mu)(b-d)\ .$$ It follows that $${\rm area}[WXYZ]=(1-2\mu)^2\ {\rm area}[ABCD]\ ,$$ and this holds for any $\mu\in[0,{1\over2}[\ $.