I have I think a solution to your question, but it might not be the most straightforward approach.
Start with the given equation
$$ \tfrac{\rm d}{{\rm d}t} \boldsymbol{p} = \boldsymbol{a} \times \boldsymbol{p} \tag{1}$$
Now decompose the location of the particle into a part parallel to $\boldsymbol{a}$ and a part perpendicular
$$ \boldsymbol{p}=\boldsymbol{a}\lambda+\left(\boldsymbol{a}\times\boldsymbol{u}\right)\mu \tag{2}$$
where $\lambda$ and $\mu$ are scalars (that can be variable at this point) and $\boldsymbol{u}$ is an arbitrary vector that is perpendicular (and thus $\boldsymbol{a} \cdot \boldsymbol{u} = 0$ at all times). This vector also might change with time at this point, but it will maintain its perpendicularity. Also note that without loss of generality, you can state that $\boldsymbol{u} \cdot \boldsymbol{u} = 1$ since the magnitude of $\boldsymbol{u}$ can be absorbed into $\mu$.
Based on the above conditions, the derivative of the position is
$$ \dot{\boldsymbol{p}}=\boldsymbol{a}\dot{\lambda}+\left(\boldsymbol{a}\times\dot{\boldsymbol{u}}\right)\mu+\left(\boldsymbol{a}\times\boldsymbol{u}\right)\dot{\mu} \tag{3}$$
where $\dot{\lambda}$, $\dot{\mu}$ and $\dot{\boldsymbol{u}}$ any changes to these quantities possible. Equate (1) to (3) to get
$$\boldsymbol{a}\dot{\lambda}+\left(\boldsymbol{a}\times\boldsymbol{u}\right)\dot{\mu}+\left(\boldsymbol{a}\times\dot{\boldsymbol{u}}\right)\mu=\boldsymbol{a}\times\left(\boldsymbol{a}\times\boldsymbol{u}\right)\mu \tag{4}$$
for which you can surmise that $\dot{\lambda}=0$ and $\dot{\mu}=0$ by matching terms left to right and (4) now takes the form
$$\mu(\boldsymbol{a}\times\dot{\boldsymbol{u}})=\mu(\boldsymbol{a}\times\left(\boldsymbol{a}\times\boldsymbol{u}\right))$$
or more simply
$$\boldsymbol{a}\times\dot{\boldsymbol{u}} =-\boldsymbol{u}\left(\boldsymbol{a}\cdot\boldsymbol{a}\right) \tag{5}$$
Note that (2) now is a planar equation because $$\require{cancel} \left(\boldsymbol{a}\cdot\boldsymbol{p}\right)=\left(\boldsymbol{a}\cdot\boldsymbol{a}\right)\lambda+\cancel{\boldsymbol{a}\cdot\left(\boldsymbol{a}\times\boldsymbol{u}\right)}\mu = \text{(const.)}$$
So we have shown that the trajectory remains on the plane perpendicular to $\boldsymbol{a}$.
Now is the time to understand the behavior of $\boldsymbol{u}$.
But before we proceed, note the initial position of the particle leads to $$\boldsymbol{p}_{0}=\boldsymbol{a}\lambda+\left(\boldsymbol{a}\times\boldsymbol{u}_{0}\right)\mu \tag{6}$$
that has the solution of
$$\begin{aligned}
\lambda & =\frac{\left(\boldsymbol{a}\cdot\boldsymbol{p}_{0}\right)}{\left(\boldsymbol{a}\cdot\boldsymbol{a}\right)} \\
\mu & = \frac{\boldsymbol{a}\cdot\left(\boldsymbol{u}_{0}\times\boldsymbol{p}_{0}\right)}{\left(\boldsymbol{a}\cdot\boldsymbol{a}\right)} \end{aligned} \tag{7}$$
and it depends on the choice of arbitrary vector $\boldsymbol{u}_0 = \boldsymbol{u}(t=0)$.
Now back to $\boldsymbol{u}$. Suppose there are two basis vectors, $\boldsymbol{e}_1$ and $\boldsymbol{e}_2$, that define $\boldsymbol{u}$ to be of unit length as such:
$$\boldsymbol{u}=\boldsymbol{e}_{1}\cos\theta+\boldsymbol{e}_{2}\sin\theta \tag{8}$$
where $\theta$ is an arbitrary value, and $\boldsymbol{e}_{1}\cdot\boldsymbol{e}_{2} = 0$. At this point, the basis vectors might change with time but we suspect they don't. We can investigate this by examining equation (5) and recognizing that
$$\dot{\boldsymbol{u}}=\dot{\boldsymbol{e}}_{1}\cos\theta+\dot{\boldsymbol{e}}_{2}\sin\theta-\boldsymbol{e}_{1}\dot{\theta}\sin\theta+\boldsymbol{e}_{2}\dot{\theta}\cos\theta \tag{9}$$
So equation (5) is now
$$\small \boldsymbol{a}\times\dot{\boldsymbol{e}}_{1}\cos\theta+\boldsymbol{a}\times\dot{\boldsymbol{e}}_{2}\sin\theta-\boldsymbol{a}\times\boldsymbol{e}_{1}\dot{\theta}\sin\theta+\boldsymbol{a}\times\boldsymbol{e}_{2}\dot{\theta}\cos\theta=-\left(\boldsymbol{a}\cdot\boldsymbol{a}\right)\left(\boldsymbol{e}_{1}\cos\theta+\boldsymbol{e}_{2}\sin\theta\right)$$
which yields two equations considering the arbitrary nature of $\theta$
$$ \begin{aligned}
(\boldsymbol{a}\times\dot{\boldsymbol{e}}_{1})+(\boldsymbol{a}\times\boldsymbol{e}_{2})\dot{\theta}+\left(\boldsymbol{a}\cdot\boldsymbol{a}\right)\boldsymbol{e}_{1}&=0\\(\boldsymbol{a}\times\dot{\boldsymbol{e}}_{2})-(\boldsymbol{a}\times\boldsymbol{e}_{1})\dot{\theta}+\left(\boldsymbol{a}\cdot\boldsymbol{a}\right)\boldsymbol{e}_{2}&=0\end{aligned}$$
Now we project the above two equations along the basis vectors, to get
$$\boldsymbol{e}_{2}\cdot\left(\boldsymbol{a}\times\dot{\boldsymbol{e}}_{1}\right)=\boldsymbol{a}\cdot\left(\dot{\boldsymbol{e}}_{1}\times\boldsymbol{e}_{2}\right)=0\\\boldsymbol{e}_{1}\cdot\left(\boldsymbol{a}\times\dot{\boldsymbol{e}}_{2}\right)=\boldsymbol{a}\cdot\left(\dot{\boldsymbol{e}}_{2}\times\boldsymbol{e}_{1}\right)=0$$
with the simple solution of
$$\dot{\boldsymbol{e}}_{1}=\omega_{2}\boldsymbol{e}_{2}\\\dot{\boldsymbol{e}}_{2}=\omega_{1}\boldsymbol{e}_{1}$$
but, it can be shown the solution of the above general solution is actually trivial.
$$\dot{\boldsymbol{e}}_{1}=0\\\dot{\boldsymbol{e}}_{2}=0$$
There is a long part here where I solve the above as an ODE using eigenvalue diagonalization but it really goes nowhere else other than $\omega_1=0$ and $\omega_2=0$ is a solution.
The result is that the basis vectors can be constant and provide a solution to (1) in the form of
$$\boxed{\boldsymbol{p}=\boldsymbol{a}\lambda+\left(\boldsymbol{a}\times\left(\boldsymbol{e}_{1}\cos\theta+\boldsymbol{e}_{2}\sin\theta\right)\right)\mu} \tag{10}$$
the above is a circle on the plane, and if we decide that $\theta=0$ when $t-0$ then
$$ \boldsymbol{e}_1 = \boldsymbol{u}_0 $$ and $\boldsymbol{e}_2$ can be derived as perpendicular to both $\boldsymbol{a}$ and $\boldsymbol{e}_1$.
I do not feel very satisfied with the above, but it gets to where you need to go (maybe). Maybe it is circular argument
