int c = a*b;
This statement actually works, since T is bounded by Integer, so after erasure, the types of a and b are Integer and they are unboxed to int.
return c;
This doesn't work since the return type of the method is not Integer, it is T, and even though <T extends Integer> and Integer is final, so T can only be Integer, the compiler doesn't allow that, since it doesn't take the finality of the type bound into account (i.e. as far as it's concerned, the method can accept instances of a sub-class of the type bound, and it can't auto-box int to any sub-class of the type bound).
Changing the return type to Integer will make the code pass compilation :
public static <T extends Integer> Integer addNumber(T a , T b){
int c = a*b;
System.out.println(c);
return c;
}
Of course, it doesn't make sense to use Integer (or any final class) as a type bound for a generic type parameter.