Given that $,\dfrac{p}{q} + \dfrac{q}{p} = 3$
Now, $\left(\dfrac{p}{q} + \dfrac{q}{p} \right)^{2} = 3^{3}$
$\implies \dfrac{p^{2}}{q^{2}} + \dfrac{q^{2}}{p^{2}} + 2 \dfrac{p}{q} \cdot \dfrac{q}{p}= 9 \quad [\because (a+b)^{2} = a^{2} + b^{2} + 2ab]$
$\implies \dfrac{p^{2}}{q^{2}} + \dfrac{q^{2}}{p^{2}} + 2 = 9$
$\implies \dfrac{p^{2}}{q^{2}} + \dfrac{q^{2}}{p^{2}} = 7$
So, the correct answer is $(B).$