Is this based on the voltage from the data sheet whereas 250v the datasheet states 1500 ohms and 200v states 1800? So this is a given based on voltage?
Bias is used to set an operating point, and to give desired operation to meet a particular goal or accommodate existing conditions.
Don't overlook the fact that resistance-coupled charts on some data sheets give conditions with a given supply voltage, and what happens to Eo (size of the output voltage swing) and Gain with various plate and cathode resistors. These charts are there to give a builder quick reference info to find suitable parts to use when they have a given supply voltage and either need maximum gain (or a particular amount of gain less than maximum), or a specific size of voltage output.
Why is 90v and 180v so common on data sheets? Because at one time all supply voltages were derived from batteries, or equipment had to be designed to operate the same way with either batteries and a.c. (or d.c.) wall voltage. 90v is ten 9v batteries, although high voltages used to be supplied by a dedicated single battery (maybe a 250v battery for your output tubes, then feeding the rest of the supply).
Anyway, short answer is note the value of bias for V1, given on the schematic as the 1.3v at the cathode. That's big enough that no pickup of the day would drive the grid appreciably close to 0v at any time, and typical guitar levels are small enough not to result in significant distortion in the output of that stage.
Where in the princeton, 240v is published on the schematic so 1k5 is used where voltage is increased the cathode resistance drops. ...
So let's draw the loadline and operating point.
See the curves below. There is a 100k loadline drawn with a 240v supply voltage; the other end of the line lands at 0v and 2.4mA (240v/100kΩ).
There are 3 cathode resistor lines drawn, for 820Ω, 1.5kΩ and 2.7kΩ. These values were chosen because they are typical 10% resistor values that were used in lots of guitar amps.
The lines are drawn by assuming a bias voltage to the left/below the loadline (0.5v for instance), and dividing the chosen voltage by the resistor value to determine the current at that point. Then a second bias point is chosen that will likely be above/right of the loadline, and the process repeated to find a second point for the cathode line.
Where the cathode resistor line crosses the loadline is the idle plate voltage and idle plate current of the stage. The location of that point in relation to the grid voltage curves also indicates the resulting bias voltage.
With the 1.5k line I drew, I get a plate voltage of ~155v and a bias voltage between -1.25 to -1.3v (a little hard for me to guess). I note that the portion of the loadline from the 0v gridline down to 0mA and 240v is nearly-bisected by the cathode line. That means the tube is biased to an operating point where the plate voltage is sitting about mid-way of the usable plate voltage swing and thereby maximizes the output signal for that tube.
Note the 820Ω line crosses the loadline further towards the 0v gridline, and the 2.7kΩ line crosses closer to cutoff at the plate voltage axis.
The 2.7k line will likely result in less output voltage swing as it will be limited during negative-going inputs by approaching cutoff at the bottom of the graph. The 2.7k cathode resistor will likely result in more distortion that the 1.5k resistor, as the internal plate resistance changes significantly during signal swing, and the spacing of the grid lines is very different during positive and negative portions of the input signal.
The 820Ω line will also result in less output voltage, as the signal swing is limited on the positive-going portion of the input signal. However, it might actually have
less distortion for small signals than the 1.5kΩ line, because the spacing of grid curves is more consistent, as is the angle of the grid lines where they cross the loadline (this indicates internal plate resistance).
So 1.5kΩ was likely chosen as the best compromise of biggest output voltage while reducing distortion to an acceptable amount. To be fair, the input signal is small enough under normal circumstances that distortion will be minimal.