Step 1: What a matter wave is.
Louis de Broglie proposed that a particle of momentum \(p\) behaves like a wave of wavelength \(\lambda=h/p\). Such a wave tied to a moving mass is a matter wave; for an electron \(\lambda=h/(m_ev)\).
Step 2: Common momentum.
Because the photon and the electron are stated to have the same \(\lambda\), and \(p=h/\lambda\), they carry an identical momentum \(p\).
Step 3: Express both energies through the same \(p\).
Photon (massless): \(E_{ph}=pc\).
Electron (massive): kinetic energy \(E_e=\dfrac{p^2}{2m_e}\).
Step 4: Take the ratio.
\[\frac{E_{ph}}{E_e}=\frac{pc}{p^2/2m_e}=\frac{2m_ec}{p}.\]
This exceeds 1 whenever \(p<2m_ec\), i.e. whenever \(\lambda>\dfrac{h}{2m_ec}\approx1.2\times10^{-12}\ \text{m}\). Real matter-wave wavelengths are always much larger than this, so the condition always holds.
Step 5: Conclusion.
For equal de Broglie wavelength, the photon's energy is larger than the electron's kinetic energy.
\[\boxed{\text{Photon has the greater energy}}\]