$$\alpha$$ and $$\beta$$ are the roots of the quadratic equation $${x^2} - x - 4 = 0$$.

$$\therefore$$ $$\alpha$$ and $$\beta$$ are satisfy the given equation.

$${\alpha ^2} - \alpha - 4 = 0$$

$$ \Rightarrow {\alpha ^{n + 1}} - {\alpha ^n} - 4{\alpha ^{n - 1}} = 0$$ ...... (1)

and $${\beta ^2} - \beta - 4 = 0$$

$$ \Rightarrow {\beta ^{n + 1}} - {\beta ^n} - 4{\beta ^{n - 1}} = 0$$ ...... (2)

Subtracting (2) from (1), we get,

$$({\alpha ^{n + 1}} - {\beta ^{n + 1}}) - ({\alpha ^n} - {\beta ^n}) - 4({\alpha ^{n - 1}} - {\beta ^{n - 1}}) = 0$$

$$ \Rightarrow {P_{n + 1}} - {P_n} - 4{P_{n - 1}} = 0$$

$$ \Rightarrow {P_{n + 1}} = {P_n} + 4{P_{n - 1}}$$

$$ \Rightarrow {P_{n + 1}} - {P_n} = 4{P_{n - 1}}$$

For $$n = 14$$, $${P_{15}} - {P_{14}} = 4{P_{13}}$$

For $$n = 15$$, $${P_{16}} - {P_{15}} = 4{P_{14}}$$

Now, $${{{P_{15}}{P_{16}} - {P_{14}}{P_{16}} - P_{15}^2 + {P_{14}}{P_{15}}} \over {{P_{13}}{P_{14}}}}$$

$$ = {{{P_{16}}({P_{15}} - {P_{14}}) - {P_{15}}({P_{15}} - {P_{14}})} \over {{P_{13}}{P_{14}}}}$$

$$ = {{({P_{15}} - {P_{14}})({P_{16}} - {P_{15}})} \over {{P_{13}}{P_{14}}}}$$

$$ = {{(4{P_{13}})(4{P_{14}})} \over {{P_{13}}{P_{14}}}}$$

$$ = 16$$