The correct option is (b) 1/[(R1 || ro) * (Ccs + Cµ)]
Explanation: The output pole is calculated, approximately, by the inverse product of the total resistance and the capacitance connected at the output node. We find that the total resistance connected to the output node is R1 in parallel with ro, due to early effect, while the total capacitance is C2 ie Ccs + Cµ. Thus, the correct option is 1/[(R1 || ro) * (Ccs + Cµ)].