Correct answer is : B

"|" is binary OR operator.
So 0x78 converts to 1111000 and 0x23 to 100011 and the bit wise OR operator is performed.

1111000 OR 100011

(1 OR 1) = 1
(1 OR 0) = 1
(1 OR 0) = 1
(1 OR 0) = 1
(0 OR 0) = 0
(0 OR 1) = 1
(0 OR 1) = 1

The above results in 1111011 and prints 123 on console