b
As an iron rod is inserted, inductance of inductor will increase.

Hence impedance \(z=\sqrt{R^{2}+(\omega L)^{2}},\) will increase,

so from \(i=\frac{v}{z}\) current will decrease and from \(P =\frac{ V _{0}^{2} R }{2 z ^{2}}\) power of bulb will decrease.