Why does Feynman say the color of a photon is a property of the source?

Note:I am new to Path Integrals/Lagrangian QFT.
In the following link, Feynman says that the color of a photon is a property of the source and that "Energy is already contained (in his Feynman diagram formulation and) there are no other things to consider". (implying no need to keep track of energy via the photon)
https://www.youtube.com/watch?time_continue=6629&v=rZvgGekvHes&feature=emb_title

I understand that the energy of a photon is related to the frequency(v) of the source by E=hv.
My question is: In Feynman’s path integral formulation, should I think of the photon as carrying energy or is this a property of the starting and ending conditions in the Lagrangian and not a property of the photon? If energy is carried by the photon (as I think it must be as I am burned ultraviolet but not by visible light even though there is more of the latter), how is information about Energy carried by the photon, for example in the Lagrangian(s) – is the energy of the photon contained in the vector potential term A, dependent on the frequency of the source? I realize that Time and Energy have a relationship via the Laplace transform but I am not sure if that helps.