Three fair dice are rolled. Given that the sum of the three numbers is a multiple of 4, the probability that their product is divisible by 4 is p/q…

Count outcomes among the $6^3=216$ ordered triples. The sums that are multiples of $4$ are $4,8,12,16$. The number of ordered triples with these sums is $3+21+25+6=55$ (for sum $4$: permutations of $(1,1,2)$; for sum $8$: $21$; for sum $12$: $25$; for sum $16$: $6$). Among them, the triples whose product is divisible by $4$ number $0+9+13+3=25$ for the sums $4,8,12,16$ respectively; the others have product either odd or twice an odd number. Hence the conditional probability is $\dfrac{25}{55}=\dfrac5{11}$. Since $\gcd(5,11)=1$, $p+q=5+11=16$. A direct enumeration over all $216$ outcomes agrees with these counts.