Correct answer is (a) \(\frac {p_2}{p_1} =(\frac {T_2}{T_1} )^{\frac {\gamma }{\gamma -1}}\)
To elaborate: The isentropic relation means no heat exchange and no dissipative forces. Also, for the calorically perfect gas, specific heats are constant. When we put these conditions in the entropy equation, we obtain the relation between pressure and temperature which is \(\frac {p_2}{p_1} =(\frac {T_2}{T_1} )^{\frac {\gamma }{\gamma -1}}\).