A parallel plate capacitor of area A and separation ‘d’ is charged to potential difference V and removed from the charging source. A dielectric slab of constant K = 3,  thickness ‘d’ and area $\frac{A}{2}$  is inserted, as shown in the figure. Let ${\sigma }_1$  be free charge density at the conductor–dielectric surface and ${\sigma }_2$ be the charge density at the conductor–vacuum surface.

Let  $\text{C}=\frac{{\text{Aε}}_{\text{0}}}{\text{d}}$
Let V be the voltage of the source. Initial charge on the capacitor $=\mathrm{CV}$
After the insertion of the dielectric, since the capacitors are attached in parallel, the potential difference across the capacitors is same and so is the electric field inside the capacitors (Ed=V’). 
Option (D) is correct.
Let $\mathrm{q}$  be the charge on the conductor-vacuum surface and $\left(\mathrm{CV}-\mathrm{q}\right)$ be the charge on the conductor-dielectric surface. 
 $\frac{\mathrm{q}}{\left(\frac{\mathrm{A}}{2}\frac{{\mathrm{\epsilon }}_0}{\mathrm{d}}\right)}=\frac{\mathrm{CV}-\mathrm{q}}{\left(\frac{\mathrm{A}}{2}\frac{{\mathrm{K\epsilon }}_0}{\mathrm{d}}\right)}\Rightarrow \mathrm{q}=\frac{\mathrm{CV}}{\mathrm{K}+1}=\frac{\mathrm{CV}}{4}$
New capacitance,  $\mathrm{C}\text{'}=\frac{{\mathrm{A\epsilon }}_0}{2\mathrm{d}}+\frac{{\mathrm{AK\epsilon }}_0}{2\mathrm{d}}=\frac{{\mathrm{A\epsilon }}_0\left(\mathrm{K}+1\right)}{2\mathrm{d}}=\frac{2{\mathrm{A\epsilon }}_0}{\mathrm{d}}=2\mathrm{C}$
Option (B) is correct.
New potential difference, $\mathrm{V}\text{'}=\frac{\mathrm{q}}{\left(\frac{{\mathrm{A\epsilon }}_0}{2\mathrm{d}}\right)}=\frac{\left(\frac{\mathrm{CV}}{4}\right)}{\left(\frac{\mathrm{C}}{2}\right)}=\frac{\mathrm{CV}}{4}\times \frac{2}{\mathrm{C}}=\frac{\mathrm{V}}{2}$
Option (C) is correct.

$\frac{{\mathrm{\sigma }}_1}{{\mathrm{\sigma }}_2}=\frac{\left(\mathrm{CV}-\mathrm{q}\right)/\left(\mathrm{A}/2\right)}{\mathrm{q}/\left(\mathrm{A}/2\right)}=\frac{\left(\mathrm{CV}-\mathrm{q}\right)}{\mathrm{q}}=\frac{\mathrm{CV}-\mathrm{CV}/4}{\mathrm{CV}/4}=3$

Option (A) is correct.