Q.A parallel plate capacitor, already charged and then disconnected from the battery, has a dielectric slab of dielectric constant $K$ inserted so as to completely fill the gap between its plates. Explain, with reasoning, what happens to

Since the capacitor is disconnected from the battery before the dielectric is inserted, there is no external path for charge to flow away -- so the charge $Q$ on the plates stays exactly the same as before the dielectric was inserted.

Capacitance increases, from Section 2.13, to $C = KC_0$, since inserting a dielectric of constant $K$ (filling the entire gap) always multiplies the vacuum/air capacitance by $K$, regardless of what is or is not connected to the plates. …

Recognise that an isolated capacitor keeps its charge fixed; use $C=KC_0$ and …

• Assuming the potential difference stays the same as before -- it does NOT, because the capacitor is isolated (charge fixed), not connected to a fixed-voltage source. …