Here is a part of the description about the circuit:
Due to the large supply inductor, only DC current flows through the inductor,
and consequently, the signal current is just the signal component of the drain current.
The drain voltage is the sum of the signal voltage and DC voltage and as the inductor is
short-circuited for DC frequencies, the DC drain voltage is the supply voltage.
It means that the peak drain voltage is 2VDD with a peak drain current of 2VDD/R.
Where is the resistor R the author is talking about? I don't see it anywhere in the circuit. I guess it is the resistance between D and S of the transistor. Do you agree?