The method on the right describes what happens if you have a balanced circuit with a balanced (differential) input signal.
The top right capacitors with the ground in the middle directly represent the input capacitances of the PMOSFETs. Since the middle node is grounded (actually Vcc but for small signal that is ground) the voltage on the node is zero, as it is grounded DUH !
Yes but as the input signal is differential, the voltage at the middle node would still be zero even if you did not ground the node. Why is that ? Because the input signal is differential, if the voltage on the left input increases x Volt then the voltage on the right input decreases x Volt.
And because the two capacitors have the same value, the voltage in the middle will not change. For small signal it (behaves as if it) is grounded !
If the capacitors confuse you, just imagine if the capacitors were resistors, the same principle will still hold.
The 3rd picture on the right shows that you can replace the 2 series capacitors as one of half their value. This gives you the differential input capacitance.
The 4th picture with the ground on one side wants to explain what the input capacitance is when applying a single ended input signal.