How is the voltage supply and peak-to-peak voltage of a rectifier found in this example?
With his calculations, you have the peak voltage already given (20) and the diode drop (0.7) so to get the 18.6V, it is just 20 - (2*0.7). That is shown in his workings. In a full wave rectifier, only 2 of the diodes are conducting at any time, which is why you use 2*0.7, which gives us the result of 18.6V.

You are correct in saying that to find a peak voltage, you can take the RMS and times by 1.414, but in this case it isn't needed as Vpeak is already given (pre-rectifier). This is the first method of finding peak voltage from a sine wave, so that may be where you were confused. But if you are given an RMS voltage in another question, you can indeed use that method to find the peak. 

The only thing he hasn't shown is how he got the 50Hz. If you have 20sin(100*pi*t), that is also the same as 20sin(wt), and w=2*pi*f. 

Knowing this, we can say 100 is 2*f. 100/2 = 50Hz

If you were unsure where the 100Hz comes from after the rectification, take a look at this graph:



This is because when the voltage is rectified, the 'negative' side of the wave is 'rectified', resulting in what is essentially a waveform with double the frequency.