If the roots of the equation $p\left(x\right)=0$ are purely imaginary then they are of the form $i\beta$.

Hence, $p\left(x\right)=a{x}^2+b$ with $a,b$ of same sign and $a,b\in R$.

Now,

$p\left(p\left(x\right)\right)=a{\left(a{x}^2+b\right)}^2+b$

If $x\in R$ or $x$ is purely imaginary, so

$x^2\in R$

$\Rightarrow p\left(x\right)\in R$

$\Rightarrow p\left(p\left(x\right)\right)\in 0$

Hence, real or purely imaginary number can not satisfy $p\left(p\left(x\right)\right)=0$, so roots cannot be rational as well. Therefore, roots of $p\left(p\left(x\right)\right)=0$ are of the form $a+ib$ where $a, b\ne 0$.