There are two possible to ways to do it :
(A*B)*C or A*(B*C)
In first case T1 = mnp + mpq = mp(n+q)
In second case T2 = npq + mnq = nq(m+p)
Let q = x , p = x+1 , n = x+2 , m = x+3
After substitution
T1 = (x+3)(x+1)(x+2+x) = 2(x+3)(x+1)(x+1)
T2 = (x+2)x(x+3+x+1) = 2(x+2)x(x+2)
T1 - T2 = 2x^2 + 6x + 3
as x >= 1 so T1-T2 > 0
hence T1 > T2
T2 requires less multiplications so the answer is c)