B O O K ==4 letters.
Your first method is wrong NOT the second! Take the two "OO" as ONE letter and you have:
3^4 ==81 permutations. The 2nd method is correct.
So, the 81 permutations look like this:
{{B, B, B, B}, {B, B, B, K}, {B, B, B, O}, {B, B, K, B}, {B, B, K, K}, {B, B, K, O}, {B, B, O, B}, {B, B, O, K}, {B, B, O, O}, {B, K, B, B}, {B, K, B, K}, {B, K, B, O}, {B, K, K, B}, {B, K, K, K}, {B, K, K, O}, {B, K, O, B}, {B, K, O, K}, {B, K, O, O}, {B, O, B, B}, {B, O, B, K}, {B, O, B, O}, {B, O, K, B}, {B, O, K, K}, {B, O, K, O}, {B, O, O, B}, {B, O, O, K}, {B, O, O, O}, {K, B, B, B}, {K, B, B, K}, {K, B, B, O}, {K, B, K, B}, {K, B, K, K}, {K, B, K, O}, {K, B, O, B}, {K, B, O, K}, {K, B, O, O}, {K, K, B, B}, {K, K, B, K}, {K, K, B, O}, {K, K, K, B}, {K, K, K, K}, {K, K, K, O}, {K, K, O, B}, {K, K, O, K}, {K, K, O, O}, {K, O, B, B}, {K, O, B, K}, {K, O, B, O}, {K, O, K, B}, {K, O, K, K}, {K, O, K, O}, {K, O, O, B}, {K, O, O, K}, {K, O, O, O}, {O, B, B, B}, {O, B, B, K}, {O, B, B, O}, {O, B, K, B}, {O, B, K, K}, {O, B, K, O}, {O, B, O, B}, {O, B, O, K}, {O, B, O, O}, {O, K, B, B}, {O, K, B, K}, {O, K, B, O}, {O, K, K, B}, {O, K, K, K}, {O, K, K, O}, {O, K, O, B}, {O, K, O, K}, {O, K, O, O}, {O, O, B, B}, {O, O, B, K}, {O, O, B, O}, {O, O, K, B}, {O, O, K, K}, {O, O, K, O}, {O, O, O, B}, {O, O, O, K}, {O, O, O, O}}==81 permutation.