$B$ cannot be derived using any functional dependency. So, it will necessarily present in candidate key.
$(AB)^+ = \big \{A,B,C,D,E\big \}$
$(BC)^+ = \big \{ B,C,D,E,A \big \}$
$(BD)^+ = \big \{ B,D,A,C,E \big \}$
$(BE)^+ = \big \{B,E \big\}$
So, above relation has $3$ candidate keys $\color{maroon}{\big \{AB,BC,BD \big \}}$
Also the given relation in "3NF"