Given : AD is median of $\Delta$ ABC
$\therefore BD=DC$ 
 ar(ΔADP) : ar(ΔABD) = 2 : 3
To find: ar(ΔPDC) : ar(ΔABC)
Construction:  Draw XY $\parallel$ BC
Median divides the triangle into two equal areas and
Triangle ABD and ADC have equal base BD and CD and are within the same parallels XY and BC.

$\therefore$ area $\Delta$ ABD = area $\Delta$ ADC...(i)
area $\Delta$ ABD : area $\Delta$ ABC = 1 : 2 ...(ii)

area $\Delta$ ADP : area $\Delta$ ABD = 2 : 3 … (iii)
area $\Delta$ ADC = area $\Delta$ ADP + area $\Delta$ PDC
area $\Delta$ ABD = area $\Delta$ ADP + area $\Delta$ PDC
area $\Delta$ PDC
= area $\Delta$ ABD - area $\Delta$ ADP
= area $\Delta$ ABD - $\frac{2}{3}$area $\Delta$ ABD
=$\frac{1}{3}$ area $\Delta$ ABD
$\therefore$area $\Delta$ PDC : area $\Delta$ ABD = 1 : 3...(iv)

$\frac{area\Delta PDC}{area\Delta ABC}=\frac{1}{3}\times \frac{1}{2}$ ….. (from equations (i) and (iv)

area $\Delta$ PDC : area $\Delta$ ABC = 1 : 6