Why does a Pull down resistor not affect AC
Based on my understanding - A popping noise is usually caused when there is some leakage from the Capacitor.


If the left side of the capacitor is not connected to anything, the capacitor can slowly discharge by itself. When it's discharged, that means the left side and the right side have the same voltage. In this case the left side will slowly go towards 4.5 volts.


  When the circuit is activated - the leftover leakage is sent to the resistor rather than the circuit, therefore avoiding a "pop" coming out of the your pedal. This makes sense to me.


Not really - when the circuit is deactivated, the tiny amount of current that leaks through the capacitor will flow through the pulldown resistor, keeping the left side at 0 volts and stopping it from going up to 4.5 volts. I must stress that leakage is a tiny amount of current. That's why a 1M resistor isn't too big.


  However, my question is more based around when the circuit it is activated. Why does the pull down resistor not affect the guitar signal?


Because it's such a big resistance, not much current flows through it. It only has a tiny effect.


  Every description I've read about a pull-down resistor implies it has no affect when the circuit is switched ON. Suggesting it's not drawing current at all?


This is incorrect, it draws current like any other resistor. But as I just said, it's a big resistor (not size, I mean it has a lot of resistance), so it draws hardly any current.