The problem asks us to find the value of $p + q^2$ given $f'(\beta) = -p/q$.

First, let's derive the expression for $f(\theta)$. We have points $A(5,0)$ and $B(-5,0)$, so $AB = 10$. In $\triangle ABC$, $AC = 6$ and $\angle A = \theta$. Point $P$ lies on segment $AC$. Let $AP = x$. Then $CP = 6 - x$. Condition (iii) states $BP + CP = 8$. Substituting $CP$, we get $BP + (6 - x) = 8 \implies BP = x + 2$. In $\triangle ABP$, applying the Law of Cosines on side $BP$: $BP^2 = AB^2 + AP^2 - 2(AB)(AP)\cos\theta$ $(x+2)^2 = 10^2 + x^2 - 2(10)(x)\cos\theta$ $x^2 + 4x + 4 = 100 + x^2 - 20x\cos\theta$ $4x + 20x\cos\theta = 96$ $x(1 + 5\cos\theta) = 24$ $x = \frac{24}{1 + 5\cos\theta}$

The area of $\triangle BCP$, denoted $f(\theta)$, can be related to the area of $\triangle ABC$. Area($\triangle ABC$) = $\frac{1}{2} AB \cdot AC \sin\theta = \frac{1}{2}(10)(6)\sin\theta = 30\sin\theta$. Since $P$ lies on $AC$, the height of $\triangle BCP$ with respect to base $CP$ is the same as the height of $\triangle ABC$ with respect to base $AC$. However, it's easier to use the ratio of bases on the same line with the same vertex $B$: $\frac{\text{Area}(\triangle BCP)}{\text{Area}(\triangle ABC)} = \frac{CP}{AC} = \frac{6-x}{6}$. So, $f(\theta) = \frac{6-x}{6} \cdot 30\sin\theta = 5(6-x)\sin\theta$. Substituting $x$: $f(\theta) = 5\left(6 - \frac{24}{1 + 5\cos\theta}\right)\sin\theta = 30\sin\theta - \frac{120\sin\theta}{1 + 5\cos\theta}$.

We need to compute $f'(\beta)$ where $\cos\beta = 4/5$ and $0 < \beta < \pi/2$. This implies $\sin\beta = 3/5$.

I will use Python to compute the derivative and evaluate it.

[PYTHON] from sympy import symbols, diff, cos, sin, S

theta = symbols('theta')

Define x as a function of theta

x = 24 / (1 + 5 * cos(theta))

Define f(theta)

Area of ABC is 30*sin(theta)

Area of BCP is (CP/AC) * Area(ABC) = ((6-x)/6) * 30sin(theta) = 5(6-x)*sin(theta)

f_theta = 5 * (6 - x) * sin(theta)

Calculate derivative f'(theta)

f_prime = diff(f_theta, theta)

Evaluate at beta where cos(beta) = 4/5, sin(beta) = 3/5

We substitute cos(theta) with c and sin(theta) with s

Note: diff might introduce sin and cos terms.

Let's substitute directly.

val = f_prime.subs({cos(theta): S(4)/5, sin(theta): S(3)/5})

print(f"Value of derivative: {val}") [/PYTHON]

The derivative of the area function $f(\theta)$ with respect to $\theta$ evaluated at $\beta$ is given by the Python output as $-\frac{96}{25}$. Thus, $f'(\beta) = -\frac{96}{25}$. We are given $f'(\beta) = -\frac{p}{q}$ where $p$ and $q$ are relatively prime positive integers. Comparing the values, we have $p = 96$ and $q = 25$. The greatest common divisor of 96 and 25 is 1, so they are relatively prime. We need to find the value of $p + q^2$.

$\boxed{721}$