It sounds a bit like you're missing something from the problem description? 5V potential relative to what? It would make sense if you have the field between two conducting plates of different potential for example.
The "0V" they are talking about in the method of mirrors is really a way of saying that any inherent excess or deficit charge in the conductor vanishes (since you assume the plate has a connection to a 0V potential reservoir of vanishing resistance) hence you have a neutral overall charge distribution on the plate.
This allows you to consider the field-distribution when for example a point-charge with charge q is placed in the vicinity of the grounded plate; by the method of images the solution is equivalent to no plate but a charge of charge -q an equivalent distance behind the plate. By differentiating the potential field you can find the equivalent distribution of charge on the plate. Without reading the actual problem, the test charge if it has a specified charge really sounds like it has to be considered in the solution.
Simply saying that a plate has a 5V potential doesn't give you the information needed to calculate a charge distribution on it or a potential field strength. It doesn't even mean that the plate is non-neutral, because Volt is a relative unit. It would have made sense if there was an equivalent plate of, say, 0V in parallel. It might also be a red herring to compound the exercise, which might be a "standard" pointcharge vs. plate problem.
The Lorentz Force does apply to an electric dipole.
An electric dipole is two charges of charge $\pm q$ separated by a distance $d=p/q.$ So start with the force $\vec{F}=q(\vec{E}+\vec{v}\times\vec{B})$ if there is no magnetic field you just get $\vec{F}=q\vec{E}$ so you get two forces, one is $(-q)\vec E(\vec r,t)$ and the other is $q\vec E(\vec r+\vec p/q).$
So the total force is $q\left(\vec E(\vec r+\vec p/q)-\vec E(\vec r,t)\right)$ which equals $q d\frac{\left(\vec E(\vec r+\vec p/q)-\vec E(\vec r,t)\right)}{d}$ which is $p\frac{\left(\vec E(\vec r+\vec p/q)-\vec E(\vec r,t)\right)}{d}$ and in the limit as $d$ goes to zero but $q$ goes to infinity in such a way that $p,$ and $\vec p$ are constant you get $p\left(\hat d\cdot \vec \nabla\right)\vec E(\vec r,t)$ and this equals $p\left(\hat p\cdot \vec \nabla\right)\vec E(\vec r,t)$ or
$\left(\vec p\cdot \vec \nabla\right)\vec E(\vec r,t).$ So even that formula is just the Lorentz Force.
So now we have to add a magnetic field instead of an electric field and we need to include that the dipole's motion since the force depends on the motion of the parts. If the dipole has two charges one with velocity $\vec v_1$ and the other with velocity $\vec v_2$ we can start by looking at the force on the average velocity $\vec v=(\vec v_1+\vec v_2)/2.$ This is actually the correct total force if the parts of the dipole don't move relative to each other.
So $\vec{F}=q(\vec{E}+\vec{v}\times\vec{B})$ becomes $\vec{F}=q(\vec{v}\times\vec{B})$ so we get So the total force is $q\vec v\times\left(\vec B(\vec r+\vec p/q)-\vec B(\vec r,t)\right)$ which equals $q d\vec v\times\frac{\left(\vec B(\vec r+\vec p/q)-\vec B(\vec r,t)\right)}{d}$ which is $p\vec v\times\frac{\left(\vec B(\vec r+\vec p/q)-\vec B(\vec r,t)\right)}{d}$ and in the limit as $d$ goes to zero but $q$ goes to infinity in such a way that $p,$ and $\vec p$ are constant you get $p\vec v\times\left(\hat d\cdot \vec \nabla\right)\vec B(\vec r,t)$ and this equals $p\vec v\times\left(\hat p\cdot \vec \nabla\right)\vec B(\vec r,t)$ or $\vec v\times\left(\vec p\cdot \vec \nabla\right)\vec B(\vec r,t).$ So that term is just the magnetic force on the dipole due to the average velocity.
So each particle has a peculiar velocity. But we also get that the magnitude $p$ doesn't change. This means that $\dot{\vec p}$ is orthogonal to $\vec p$ (because $0=\frac{\mathrm d}{\mathrm d t}(\vec p\cdot \vec p)=2\vec p\cdot\dot{\vec p}$).
So what needs to be done? You can relate the peculiar motion $\vec v_1-\vec v$ and $\vec v_2-\vec v$ to $\dot{\vec p}$ and then compute the force due to that velocity and add it to the other force to get the total force.
Best Answer
It is better to calculate the potential energy of the dipole directly by the formula
$$U=-\overrightarrow{p_1}\cdot\overrightarrow{E_2}$$
where $\overrightarrow{p_1}$ represents the dipole moment of the dipole and $\overrightarrow{E_2}$ is the electric field of the image dipole $\overrightarrow{p_2}$.
Now $$\overrightarrow{E_2}=\frac{3(\overrightarrow{p_2}\cdot\overrightarrow{r})\overrightarrow{r}}{r^5}-\frac{\overrightarrow{p_2}}{r^3}$$
where $\overrightarrow{r}$ is radius vector from the image dipole to its original.
In given case the magnitude of $\overrightarrow{r}$ is $r=2L$ and $\overrightarrow{p_2}=\overrightarrow{p}$ and $\overrightarrow{p_1}=\overrightarrow{p}$ and $\overrightarrow{r}$ is parallel to $\overrightarrow{p}$
After the dot product calculation we have the potential energy of the dipole to be
$$U=-\overrightarrow{p_1}\cdot\overrightarrow{E_2}=-\frac{2p^2}{r^3}$$
Since $U<0$ the dipole and the plate attract each other.
Now, the attracting force between the dipole and plate it is obtained by differentiating of the potential energy by r $$F=-\frac{dU}{dr}=\frac{6p^2}{r^4}$$
Since $r=2L$ the magnitude of the force
$$F=\frac{3p^2}{8L^4}$$