Solution: Given that if $\lambda $ is a root of $p(x) = x^4 -4x^3 +2x^{2} +ax+b$, then $\frac{1}{\lambda }$ is also a root of $p$. So, let us suppose that the roots of the equation is $\lambda ,\frac{1}{\lambda }, \mu ,\frac{1}{\mu }$. We recall that if $p(x) = ax^4 + bx^3 + cx^2 + dx +e$ and $\alpha ,\beta \gamma $ and $\delta $ are roots of it, then \begin{align*} & \sum \alpha \coloneqq \alpha + \beta + \gamma + \delta = -\frac{b}{a}\\ & \sum \alpha \beta = \frac{c}{a}\\ & \sum \alpha \beta \delta = -\frac{d}{a},~\text{and }\\ & \alpha\beta \gamma \delta = \frac{e}{a}. \end{align*}

So, we have \begin{align*} \lambda \times \frac{1}{\lambda }\times \mu \times \frac{1}{\mu } = \frac{b}{1} & \implies b = 1. \end{align*} Similarly, \begin{align*} & \lambda \cdot \frac{1}{\lambda }\cdot \mu + \lambda \cdot \frac{1}{\lambda }\cdot\frac{1}{\mu } + \lambda \cdot \mu \cdot \frac{1}{\mu } + \frac{1}{\lambda }\cdot \mu \cdot \frac{1}{\mu } = -a \\ \implies & \mu +\frac{1}{\mu }+\lambda +\frac{1}{\lambda } = -a \\ \implies & -(-4) = -a \\ \implies & a = -4. \end{align*}