Right choice is (a) 75 Ω
The best I can explain: Hybrid parameter h11 is given by, h11 = \(\frac{V_1}{I_1}\), when V2=0.
Therefore short circuiting the terminal Y-Y’, we get,
V1 = I1 ((50||50) + 50)
= I1 \(\left(\left(\frac{50×50}{50+50}\right) + 50\right)\)
= 75I1
∴ \(\frac{V_1}{I_1}\) = 75.
Hence h11 = 75 Ω.