[THOUGHT] We have points A(5,0) and B(-5,0). Let C be such that AC = 6 and ∠A = θ.
Place C using polar coordinates from A:
C = (5 - 6 cos θ, 6 sin θ).

Let P divide AC with ratio λ (0<λ<1):
P = (5 - 6 λ cos θ, 6 λ sin θ).

Condition BP + CP = 8 gives CP = 6(1-λ) and BP = 2+6λ.
Using BP² = (10-6λ cos θ)² + (6λ sin θ)² = 100 -120λ cos θ +36λ²,
equating to (2+6λ)² yields λ = 4/(5 cos θ + 1).

Area of triangle BCP:
f(θ) = ½ |(C−B)×(P−B)| = 30 sin θ (1−λ)
= 30 sin θ · (5 cos θ−3)/(5 cos θ+1).

Derivative:
f'(θ) = 30·(25 cos³θ + 10 cos²θ − 3 cosθ − 20)/(5 cosθ+1)².

For β with cos β = 4/5:
cosβ = 4/5, sin²β = 1−(16/25)=9/25 ⇒ sinβ = 3/5.
Plug into numerator:
25·(64/125)+10·(16/25)−3·(4/5)−20 = -16/5.
Denominator: (5·4/5+1)² = 5² = 25.

Thus f'(β) = 30·(-16/5)/25 = -96/25.

So p/q = 96/25 ⇒ p = 96, q = 25 (coprime).
Required value: p + q² = 96 + 625 = 721.

We verify numerically with Python. [/THOUGHT] [PYTHON] import sympy as sp

theta = sp.symbols('theta', real=True) cos_theta = sp.cos(theta) sin_theta = sp.sin(theta)

lambda expression

lam = 4/(5*cos_theta + 1)

area function f(theta)

f = 30sin_theta(1 - lam) f_simplified = sp.simplify(f) print("f(theta) =", f_simplified)

derivative

f_prime = sp.diff(f_simplified, theta) print("f'(theta) =", sp.simplify(f_prime))

evaluate at beta where cos(beta) = 4/5

cos_beta_val = sp.Rational(4,5) sin_beta_val = sp.sqrt(1 - cos_beta_val**2) # positive beta = sp.acos(cos_beta_val) f_prime_at_beta = f_prime.subs(theta, beta).evalf() print("f'(beta) =", f_prime_at_beta)

rational form

f_prime_at_beta_rational = sp.simplify(f_prime.subs(theta, beta)) print("f'(beta) rational =", f_prime_at_beta_rational) [/PYTHON]

\boxed{721}