There are two ways of solving this problem

Method 1:
Pick 3 departments out of 4 who will provide us that one-one member in 4C3 ways = \(\frac{4!}{1! * 3!}\)
= 4 ways

Pick one one member from these departments in 4C1 * 4C1 * 4C1 = 4*4*4 = \(4^3\) ways
=> Total number of ways = 4*\(4^3\) = \(4^4\)

Method 2:

Consider the following three dashes as the people in the group

___ ___ ___

Now for the first place we have 16 choices because we can pick any member from any department

Now, once we pick the first person in 16 ways then
The second person can be picked from any department apart from the first person's department
So, we have only 12 choices for the second position
(all 4 from the first department, one who is chosen and remaining 3 from that person's department cannot be chosen)
So, second can be picked in 12 ways

Now third can be picked in 8 ways
(all 4 from the two department whose people are already picked cannot be picked now)
So, 8 ways for third person

Total Number of ways = \(\frac{(16 * 12 * 8)}{3!}\) [ Dividing by 3! as order does not matter ]

= 16 * 2* 8 = \(4^4\)

So, Answer will be B
Hope it helps!