A dielectric slab is introduced between the plates of an isolated...

For an isolated charged capacitor, the battery is disconnected. Thus, the charge on the capacitor plates has no path to escape or increase, so it remains constant.

When a dielectric slab of dielectric constant $K$ is introduced, the capacitance increases from $C$ to $KC$.

Since charge $Q$ is constant, the potential difference $V = \dfrac{Q}{C}$ becomes $V' = \dfrac{Q}{KC} = \dfrac{V}{K}$, which means it decreases.

The electric field $E = \dfrac{V}{d}$ becomes $E' = \dfrac{V'}{d} = \dfrac{E}{K}$, which means it decreases.

The stored energy $U = \dfrac{Q^2}{2C}$ becomes $U' = \dfrac{Q^2}{2KC} = \dfrac{U}{K}$, which means it decreases.

Therefore, only the charge on the capacitor remains unchanged.