a
(a) Let the piston be displaced through distance \(x\) towards left, then volume decreases, pressure increases. If \(\Delta P\) is increase in pressure and \(\Delta V\) is decrease in volume, then considering the process to take place gradually (i.e. isothermal) \({P_1}{V_1} = {P_2}{V_2}\)

\( \Rightarrow PV = (P + \Delta P)(V - \Delta V)\)

\( \Rightarrow PV = PV + \Delta PV - P\Delta V - \Delta P\Delta V\)

\( \Rightarrow \Delta P.V - P.\Delta V = 0\) (neglecting \(\Delta P.\Delta V)\)

\(\Delta P(Ah) = P(Ax)\)

\( \Rightarrow \Delta P = \frac{{P.x}}{h}\)

This excess pressure is responsible for providing the restoring force \((F)\) to the piston of mass \(M\).

Hence \(F = \Delta P.A = \frac{{PAx}}{h}\)

Comparing it with \(|F| = kx\)

\(\Rightarrow k = M{\omega ^2} = \frac{{PA}}{h}\)

\( \Rightarrow \omega = \sqrt {\frac{{PA}}{{Mh}}} \)

\(\Rightarrow T = 2\pi \sqrt {\frac{{Mh}}{{PA}}} \)

Short trick : by checking the options dimensionally. Option (a) is correct.