Given configuration

Two identical point dipoles are placed on the $x$-axis, separated by distance $a$.

$\vec p_1 = p\hat i, \qquad \vec p_2 = -p\hat i$

Their dipole moments remain fixed in direction while they move only along the $x$-axis.

Potential energy of interaction of two point dipoles

The interaction energy of two dipoles separated by vector $\vec r$ is

$U=\frac{1}{4\pi\varepsilon_0 r^3}\left[\vec p_1\cdot \vec p_2-3(\vec p_1\cdot \hat r)(\vec p_2\cdot \hat r)\right]$

Here, the dipoles lie on the same axis, so take $\hat r=\hat i$ and $r=a$.

Compute the dot products

Since

$\vec p_1=p\hat i, \qquad \vec p_2=-p\hat i$

we get

$\vec p_1\cdot \vec p_2 = -p^2$

Also,

$\vec p_1\cdot \hat r = p, \qquad \vec p_2\cdot \hat r = -p$

so

$(\vec p_1\cdot \hat r)(\vec p_2\cdot \hat r)= -p^2$

Therefore,

$U=\frac{1}{4\pi\varepsilon_0 a^3}\left[-p^2-3(-p^2)\right]$

$U=\frac{1}{4\pi\varepsilon_0 a^3}(2p^2)$

$U_i=\frac{p^2}{2\pi\varepsilon_0 a^3}$

At infinite separation

As the dipoles go infinitely far apart,

$U_f=0$

Since initially they are released from rest, initial kinetic energy is zero.

Apply conservation of energy

Initial total energy:

$E_i=U_i=\frac{p^2}{2\pi\varepsilon_0 a^3}$

Final total energy:

Each dipole has speed $v$, so total kinetic energy is

$K_f=2\left(\frac{1}{2}mv^2\right)=mv^2$

Hence,

$mv^2=\frac{p^2}{2\pi\varepsilon_0 a^3}$

$v^2=\frac{p^2}{2\pi\varepsilon_0 m a^3}$

$v=\frac{p}{a}\sqrt{\frac{1}{2\pi\varepsilon_0 m a}}$

Match with options

This matches Option C:

$\boxed{\frac{p}{a}\sqrt{\frac{1}{2\pi\varepsilon_0 m a}}}$