We'll do each case separately then we'll
put the four cases into a truth table:

(P → Q) ↔ (¬P ∨ Q)

Let "T" mean "true" and let "F" be "false:

Case 1:  P is true and Q is true:

(P → Q) ↔ (¬P ∨ Q)
(T → T) ↔ (¬T ∨ T)
   T    ↔ ( F ∨ T)
   T    ↔     T
        T

Case 2:  P is true and Q is false:

(P → Q) ↔ (¬P ∨ Q)
(T → F) ↔ (¬T ∨ F)
   F    ↔ ( F ∨ F)
   F    ↔     F
        T

Case 3:  P is false and Q is true:

(P → Q) ↔ (¬P ∨ Q)
(F → T) ↔ (¬F ∨ T)
   T    ↔ ( T ∨ T)
   T    ↔     T
        T  

Case 4:  P is false and Q is false:

(P → Q) ↔ (¬P ∨ Q)
(F → F) ↔ (¬F ∨ F)
   T    ↔ ( T ∨ F)
   T    ↔     T
        T  

It is true in all four cases, so it is a tautology.

Or you can do it as a truth table which is equivalent to
the four cases above:

        | P | Q | ¬P | P → Q | ¬P ∨ Q | (P → Q) ↔ (¬P ∨ Q |
case 1: | T | T |  F |   T   |    T   |         T          | 
case 2: | T | F |  F |   F   |    F   |         T          |  
case 3: | F | T |  T |   T   |    T   |         T          |  
case 4: | F | F |  T |   T   |    T   |         T          |    

Since the last column came out all trues, the original statement
is a tautology.

Edwin