A parallel plate capacitor has unequal charges $+2\mathrm{Q}$ and $-\mathrm{Q}$ as shown in the figure. If the capacitance of the capacitor is C and P is a point outside of the capacitor, then (d is the distance between the plates)

Charges on various faces of the plates will be as shown in the figure.

$E_P=\frac{Q}{4A{\epsilon }_0}+\frac{Q}{4A{\epsilon }_0}=\frac{Q}{2A{\epsilon }_0}\ne 0$. Hence a point charge kept at $P$ will
experience a non-zero force     Option (a) is wrong.
Electric field between the plates, $E=\frac{3Q}{2A{\epsilon }_0}$
Potential difference between the plates,  $V=Ed=\frac{3Q}{2A{\epsilon }_0}d$
But $C=\frac{A{\epsilon }_0}{d}$

$\Rightarrow V=\frac{3Q}{2C}$                      Option (b) is correct
The energy stored in the electric field in the region between the plates is
 $U=\frac{1}{2}C{V}^2=\frac{1}{2}C{\left(\frac{3Q}{2C}\right)}^2=\frac{9Q^2}{8C}$              Option (c) is correct.
The force on one plate due to the other plate is

$F=\left(\text{electric\hspace{0.17em}\hspace{0.17em}field\hspace{0.17em}\hspace{0.17em}on\hspace{0.17em}\hspace{0.17em}the\hspace{0.17em}\hspace{0.17em}plate}\right)\left(\text{charge\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}on\hspace{0.17em}\hspace{0.17em}plate}\right)$

$F=\left(\frac{E}{2}\right)\left(\frac{3Q}{2}\right)=\left(\frac{3Q}{4A{\epsilon }_0}\right)\left(\frac{3Q}{2}\right)=\frac{9Q^2}{8A{\epsilon }_0}$         Option (d) is wrong.