The schematic shows a peak detector but onl one polarity. Not peak-to-peak.
In the schematic the internal source impedance is missing. If you have 550 VAC source, then the capacitor (100 uF) will charge to a peak voltage of SQRT(2) x 550 V, or 780 V. I doubt you can find such capacitor on the market. As the capacitor would discharge by more than 1MOhm in the voltage divider on the left, I would use a 0.1 to 1.0 uF capacitor with 1 kV rating, easier to find.
With 800 V max peak, use a 3kV reverse diode, and use a transformer from AC circuit, not the ground shown, for safety.
Resistors in the voltage divider should be rated to U squared/ R, therefore 0. 64 W, the other resistors only 1/8 Watt.
As the peak detector will accept AC voltage at 50 or 60 Hz, the easier solution (also for safety), use a small AC transformer, 220 to 6 or 12 V, then you can use a low-voltage diode like 1N4001 and the capacitor rated for 25 V. Adjust the resistors in voltage divider to get a suitable voltage for your ADC. The mentioned transformer can process also 50/60 Hz harmonics, so the peak detector can accept also "abrupt" peaks coming on mains line.
Use the ground only on transformer secondary as shown in the schematic.