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(B) The energy $E$ of a photon is given by $E = \frac{hc}{\lambda}$,where $h$ is Planck's constant,$c$ is the speed of light,and $\lambda$ is the wave

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$P$ and $E$ both will increase

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(B) The energy $E$ of a photon is given by $E = \frac{hc}{\lambda}$,where $h$ is Planck's constant,$c$ is the speed of light,and $\lambda$ is the wavelength.The momentum $P$ of a photon is given by $P = \frac{h}{\lambda}$.From these relations,it is clear that both $E$ and $P$ are inversely proportional to the wavelength $\lambda$ ($E \propto \frac{1}{\lambda}$ and $P \propto \frac{1}{\lambda}$).Therefore,if the wavelength $\lambda$ is reduced,both the energy $E$ and the momentum $P$ will increase.

If $E$ and $P$ are the energy and the momentum of a photon respectively,then on reducing the wavelength of the photon:

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