Answer
$(a)$ increasing the frequency of the incident light.
Work Step by Step
As we know:
$E=$ $1/2mv^2$ $+$ $\phi$
Energy $=$ kinetic energy $+$ work function
As $E=hf$ , we know that work function is a constant so kinetic energy depends on the frequency of incident light. As frequency of the incident light increases kinetic energy also increases.
Increasing the number of photon increases the intensity of light not the kinetic energy hence option $(b)$ is incorrect.
Selecting the metal that has greater work function decreases the kinetic energy hence option $(d)$ is incorrect.
Using the photon whose frequency $f_°$ is less than$W/f_°$ where $W_°$ is the work function of the metal then the energy of the incident light is less than the work function of the metal and electron does not eject from the surface of metal so kinetic energy is zero and hence option $c$ is incorrect.
