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.