Groups: (1, 3, 5, 7, 9, 11, 13, 15) → D'; (6, 7, 14, 15) → (B' + C'); (0, 1) → (A + B + C)
Reduced expression: F = D'(B' + C')(A + B + C)
Grouping the 0s: octet {1,3,5,7,9,11,13,15} = D', quad {6,7,14,15} = (B' + C'), pair {0,1} = (A + B + C).
Answer (b)
Logic gate diagram for $F=(A+B+C)\cdot(B'+C')\cdot D'$: a 3-input OR gate (A, B, C) and a 2-input OR gate (B', C') feed a 3-input AND gate together with D'.