[THOUGHT] The given hyperbola is (\dfrac{x^2}{a^2} - \dfrac{y^2}{a^2} = -1), which can be rewritten as (\dfrac{y^2}{a^2} - \dfrac{x^2}{a^2} = 1). This is a hyperbola centered at the origin opening upwards and downwards with (a = b). The foci are at ((0, \pm a\sqrt{2})).

The tangent line at ((a, \sqrt{2}a)) is derived using the hyperbola tangent formula: [ \frac{yy_0}{a^2} - \frac{xx_0}{a^2} = 1 \implies \frac{y \cdot \sqrt{2}a}{a^2} - \frac{x \cdot a}{a^2} = 1 \implies \sqrt{2}y - x = a. ] Setting (x = 0) (intersection with the y-axis) gives (y = \frac{a}{\sqrt{2}}). Thus, point (P) is ((0, \frac{a}{\sqrt{2}})).

Distances from (P) to the foci (F(0, a\sqrt{2})) and (F'(0, -a\sqrt{2})) are: [ PF = \left| a\sqrt{2} - \frac{a}{\sqrt{2}} \right| = \frac{a}{\sqrt{2}}, \quad PF' = \left| \frac{a}{\sqrt{2}} + a\sqrt{2} \right| = \frac{3a}{\sqrt{2}}. ] Their product is: [ PF \times PF' = \frac{a}{\sqrt{2}} \times \frac{3a}{\sqrt{2}} = \frac{3a^2}{2}. ] Setting this equal to 8: [ \frac{3a^2}{2} = 8 \implies a^2 = \frac{16}{3}. ] Thus, (p = 16) and (q = 3), giving (p + q = 19).

[PYTHON]

Verification of the solution

from sympy import symbols, solve, sqrt

a = symbols('a', positive=True, real=True)

Equation from the product of distances

equation = (3 * a2) / 2 - 8 solution = solve(equation, a2) a_squared = solution[0] p, q = a_squared.as_numer_denom() print(p + q) [/PYTHON]

[THOUGHT] The Python output confirms that ( p + q = 19 ) when ( a^2 = \frac{16}{3} ). This matches the derived solution. [/THOUGHT] \boxed{19}