Let the product of $$\omega_1 = (8 + i)\sin\theta + (7 + 4i)\cos\theta$$ and $$\omega_2 = (1 + 8i)\sin\theta + (4 + 7i)\cos\theta$$ be $$\alpha + i\beta$$, $$i = \sqrt{-1}$$. Let p and q be the maximum and the minimum values of $$\alpha + \beta$$ respectively.

Write the two complex numbers as
$$\omega_1 =(8+i)\sin\theta +(7+4i)\cos\theta ,\qquad \omega_2 =(1+8i)\sin\theta +(4+7i)\cos\theta .$$

Their product is
$$\omega_1\omega_2 =(8+i)\!(1+8i)\sin^2\theta +\big[(8+i)(4+7i)+(7+4i)(1+8i)\big]\sin\theta\cos\theta +(7+4i)(4+7i)\cos^2\theta .$$

Compute the four required products one by one.

1. $$AC=(8+i)(1+8i)=65i.$$

2. $$AD=(8+i)(4+7i)=25+60i.$$

3. $$BC=(7+4i)(1+8i)=-25+60i.$$

4. $$BD=(7+4i)(4+7i)=65i.$$

Hence
$$AD+BC=(25+60i)+(-25+60i)=120i.$$

Substitute back:

$$\omega_1\omega_2 =65i\sin^2\theta+120i\sin\theta\cos\theta+65i\cos^2\theta.$$

Factor $$65i$$ in the first and last terms and use $$\sin^2\theta+\cos^2\theta=1$$:

$$\omega_1\omega_2 =65i+120i\sin\theta\cos\theta.$$

Convert the mixed term with $$\sin2\theta=2\sin\theta\cos\theta$$:

$$\omega_1\omega_2 =65i+60i\sin2\theta.$$

Therefore the product is purely imaginary. Write
$$\omega_1\omega_2=\alpha+i\beta,$$ so that $$\alpha=0,\quad \beta=65+60\sin2\theta.$$

The required sum is $$\alpha+\beta=\beta=65+60\sin2\theta.$$

Because $$-1\le\sin2\theta\le1,$$ the maximum and minimum values are

$$p=65+60(1)=125,\qquad q=65+60(-1)=5.$$

The problem asks for $$p+q,$$ which is

$$p+q=125+5=130.$$

Hence the correct choice is Option B (130).