Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
According to Einstein’s theory of relativity
$$E^2-p^2c^2 =mc^2$$
So a massless photon has a mass of zero,
$$E^2-p^2c^2 =(0)c^2=0$$
Hence, its momentum is given by
$$\boxed{p =\dfrac{E}{c}}$$
$$\color{blue}{\bf [b]}$$
According to Einstein’s claimed that the energy of a photon is related to its frequency by $E=hf$
Plug that into the boxed formula above,
$$p =\dfrac{hf}{c}$$
where $f=c/\lambda$
$$p =\dfrac{h\color{red}{\bf\not} c}{\color{red}{\bf\not} c\lambda}$$
Thus,
$$\boxed{\lambda=\dfrac{h}{p}}$$
$$\color{blue}{\bf [c]}$$
According to classical physics, $p=mv$, plug that into the previous boxed formula.
$$\boxed{\lambda=\dfrac{h}{mv}}$$
We got a familiar expression which is the expression for de Broglie wavelength.
