(b) $Z = \sqrt{R^{2} + X_{C\ }^{2}}\ :I_{rms} = \frac{V_{rms}}{Z}:P = I_{rms}^{2}R$
Where $X_{C} = \frac{1}{\omega C}$
As $\omega$ is increased, $X_{C}$will decrease or$\ Z\ $will decrease$.$ Hence $I_{rms}\ or\ P\ $will increase$.$ Therefore, bulb glows brighter.
Hence the correct option is (b).