This was not likely to have been an SAT practice problem, though it is a typical contest problem.
$AE:EC = x:y$ (since those two triangles have the same altitude to $AC$, the ratios of their areas is the ratios of their bases with respect to that altitude) and $AP:PD = (x+y):z$ (same idea as $AE:EC$). Knowing these two ratios, apply the technique of mass points, putting masses $zy$ at $A, zx$ at $C$ (gives the ratio $x:y$ for $AE:EC$), and $y(x+y)$ at $D$ (gives $(x+y):z$ for $AP:PD$). This results in a mass of $y(x+y)-zx$ at $B$, so the ratio $BD:DC = zx:(y(x+y)-zx).$ This must also be the ratio of the areas of △ABD to △ADC (common altitude again), so (area of △ABD):$(x+y+z) = zx:(y(x+y)-zx)$. Solving from there is a matter of bashing out the algebra.