Ah! I see what you are asking about.
The 65N1P, at the Seduction operating point, has about 10.6K plate resistance. If you are using the C4S plate load then the output resistance of the first stage is the plate resistance. So 66.5K + (66.5K*66.5K/10.6K) is 484K ohms.
I chose to do the Seduction this way because the equations for the RIAA equalization are available, but usually based on zero source impedance (rp) and infinite load resistance (Rgrid). This way the network always has the standard values, even when you change tubes.
To analyze it (if you or anyone else wants to) look at the circuit and you will see that the 66.5K is in series with the source resistance rp, and the grid resistor goes from that point to ground. The imaginary souce itself has zero impedance to ground, plus the resistance rp, so the effective resistance is Rgrid in parallel with (rp+66.5K). With a little algebra, you can calculate Rgrid given rp - the result is what I posted.
This method makes the RIAA accuracy less dependent on an exact knowledge of rp, while still keeping the overall impedance of the network low so that it is less likely to pick up hum and noise from stray electrical fields or RF noises.
