The digits of N can be 2, 4, 6 and 8. Since N is divisible by 4, the last two digits must be 24 or 28 or 44 or 48 or 64 or 68 or 84 or 88.
When the last two digits are the same, i.e., 44 or 88, there can be no more repetition; then, the first three digits can be chosen in 3! Ways. Thus, the number of possibilities = 3! x 2 = 12.
When the last two digits are distinct, we get the following cases:
i] Third digit is same as fourth digit.
Number of ways remaining two digits can be chosen = 2!
Ii] Third digit is same as second digit.
Number of ways = 2!
iii] Second digit is same as first digit.
Number of ways = 2!
Thus, when the number ends with 24, the number of possibilities = 2! + 2! + 2! = 6.
Since there are 6 cases where the last two digits are distinct, the total number of ways =

6 $\times$ 6+ 12 = 36 + 12 = 48. Hence, [b].