Two identical solid spheres A and B of mass m and radius r each have short light identical dipoles embedded at their respective centers. B is in equilibrium on an incline of inclination $\theta$ as shown such that its dipole moment is parallel to the incline. Sphere A is released, in the position shown, on a rough horizontal surface. Both A and B are located in electric field of a uniform infinite sheet of surface charge density $\sigma$. If friction between A and horizontal surface is sufficient to prevent slipping, (neglect mutual interaction between A and B).

Consider FBD of sphere B as shown in Fig.1.

About contact point O of the sphere with the incline,
${\tau }_1=$ torque of $mg\sin \theta =mg\sin \theta .r$ (anti-clockwise)
${\tau }_1=$ torque of electric field E due to infinite sheet = $pE\sin \theta$,(clockwise)
For equilibrium of B,  ${\tau }_1={\tau }_2\Rightarrow mg\sin \theta .r=pE\sin \theta$  
$\Rightarrow mgr=pE\dots \dots \dots \dots \dots \dots \dots \dots \mathrm{..}\left(i\right)$
Consider FBD of sphere A, just after release as shown in Fig.2.
Let a and $\alpha$ are acceleration of center of mass and angular acceleration of the sphere, respectively.

Moment of inertia of the sphere about O, using parallel axes theorem,
$I=I_{CM}+m{d}^2=\frac{2}{5}m{r}^2+m{r}^2=\frac{7}{5}m{r}^2\mathrm{..................}\left(ii\right)$
Torque of electric force of sheet on dipole, $\tau =pE\sin {90}^0=pE\mathrm{..........}\left(iii\right)$ 
Using Eq.s (iii) and (ii), We get 
Angular acceleration of the sphere is $\alpha =\frac{\tau }{1}=\frac{pE}{\frac{7}{5}m{r}^2}=\frac{5pE}{7m{r}^2}$
By condition of rolling without slipping,  $a=\alpha r=\frac{5pE}{7mr}=\frac{5mgr}{7mr}$
[using Eq. (i)] $\Rightarrow a=\frac{5g}{7}\mathrm{..................}\left(iv\right)$    $\therefore$Option (a) is correct.
From FBD in Fig.2, using, F=ma   in horizontal direction, We get $f=m\left(\frac{5g}{7}\right)=\frac{5mg}{7}4$
[using Eq.(iv)] $\therefore$  Option (b) is incorrect.
Due to electric field torque experienced by sphere A, will rotate the dipole clockwise. 
Potential energy of a dipole is given by   $U=-pE\cos \theta$
Therefore,  $U_{initial}=-pE\cos {90}^0=\mathrm{0..............}\left(v\right)$
                    $U_{final}=-pE\cos 0^0=-pE\mathrm{.............}\left(vi\right)$
Considering, rotation of sphere by  $\frac{\pi }{2}$,
Gain in KE = Loss in PE $\frac{1}{2}I{\omega }^2=U_{initial}-U_{final}$
       $\begin{array}{l}\frac{1}{2}.\frac{7}{5}m{r}^2{\omega }^2=0-(-pE)\Rightarrow \omega =\sqrt{\frac{10pE}{7m{r}^2}}=\sqrt{\frac{10p\left(\frac{\sigma }{2{\in }_0}\right)}{7m{r}^2}}\\ =\sqrt{\frac{5\rho \sigma }{7m{r}^2{\in }_0}}\end{array}$  
$\Rightarrow$Option © is incorrect.
Finally, When dipole is parallel to horizontal surface, torque due to electric force is equal to    

$\tau =pE\sin 0^0=0\Rightarrow {\alpha }_{CM}=\frac{\tau }{1}=0\Rightarrow f^{|}=m{a}_{CM}=0\therefore$ Option (d) is correct.