Let the probability of getting head for a biased coin be (1)/(4) ....

Given a quadratic equation $64{x}^{2} + 5nx + 1 = 0$ has no real roots, so

$D < 0$

$⇒ 25{N}^{2} - 4 \times 64 < 0$

$⇒ N < \dfrac{16}{5} ⇒ N = 1, 2, 3$

And, For a biased coin, the probability of getting head is $P = \dfrac{1}{4},$ and probability of getting tail will be $Q = \dfrac{3}{4}$

So, required probability is

$\dfrac{1}{4} + \dfrac{3}{4} \cdot \dfrac{1}{4} + {\left(\dfrac{3}{4}\right)}^{2} \cdot \dfrac{1}{4} + {\left(\dfrac{3}{4}\right)}^{3} \cdot \dfrac{1}{4} = \dfrac{p}{q}$

$⇒ \dfrac{\dfrac{1}{4}\left(1 - {\left(\dfrac{3}{4}\right)}^{3}\right)}{1 - \left(\dfrac{3}{4}\right)} = \dfrac{p}{q}$

$⇒ 1 - {\left(\dfrac{3}{4}\right)}^{3} = \dfrac{p}{q}$

$⇒ 1 - \dfrac{27}{64} = \dfrac{p}{q}$

$⇒ \dfrac{p}{q} = \dfrac{37}{64}$

$q - p = 27$