Well, the construction of the BDD is done as usual, i.e. either you do a repeated Shannon decomposition or you work bottom-up using the Apply algorithm. Using case distinctions, you get

case a=0: b&!c&d | !b&c&d | b&c&!d

case b=0: c&d

case b=1: !c&d | c&!d

case a=1: b&!c&!d | !b&c&!d & !b&!c&d

case b=0: c&!d & !c&d

case b=1: c&!d

Hence, you need BDDs for c&d, !c&d, c&!d which are the ones you see in the figure of the solution of the exercise, and the binary tree with nodes a and b makes the above case distinctions.