Suppose the frequency and intensity of a light source are both...

The intensity $I$ of a light source is given by $I = n h \nu$, where $n$ is the number of photons incident per unit area per unit time and $\nu$ is the frequency of the light.

When both the intensity and the frequency are doubled, the new intensity is $I' = 2I$ and the new frequency is $\nu' = 2\nu$.

The new number of incident photons per unit area per unit time is $n' = \dfrac{I'}{h \nu'} = \dfrac{2I}{h(2\nu)} = \dfrac{I}{h\nu} = n$.

Since the number of incident photons remains unchanged, the number of emitted photoelectrons per unit time remains the same. Thus, the saturation photocurrent remains unchanged. Statement A is correct.

According to Einstein's photoelectric equation, the maximum kinetic energy of the photoelectrons is $K = h\nu - \Phi$, where $\Phi$ is the work function of the material.

When the frequency is doubled, the new maximum kinetic energy is $K' = h(2\nu) - \Phi = 2h\nu - \Phi = 2(K + \Phi) - \Phi = 2K + \Phi$.

Since $\Phi > 0$, we get $K' > 2K$. The maximum kinetic energy is more than doubled, not exactly doubled. Statement B is incorrect.