Answer
Yes. They are identical.
Work Step by Step
The de Broglie wavelength is given by $\lambda_{dB}=\frac{h}{p}$.
For a photon, E=pc, so the photon has momentum $p=\frac{E}{c}$.
Substituting, we see that $\lambda_{dB}=\frac{hc}{E}$.
However, a photon’s energy is equal to hf, so we have $\lambda_{dB}=\frac{hc}{hf}=\frac{c}{f}=\lambda$.
This is the wavelength of the associated electromagnetic wave.
