1. Given hyperbola and point on it

We have $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and the point $P(4,2\sqrt{3})$ lies on it.

So, $\frac{16}{a^2}-\frac{12}{b^2}=1 \qquad ...(1)$

2. Use the product of focal distances

For the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$ the foci are at $F_1(-c,0),\quad F_2(c,0),$ where $c^2=a^2+b^2.$

If the distances of $P(x,y)$ from the foci are $r_1$ and $r_2$, then $r_1r_2=\sqrt{\big((x+c)^2+y^2\big)\big((x-c)^2+y^2\big)}.$

For $P(4,2\sqrt3)$, $r_1r_2=\sqrt{\big((4+c)^2+12\big)\big((4-c)^2+12\big)}=32.$

Squaring, $\big((4+c)^2+12\big)\big((4-c)^2+12\big)=1024.$

Now, $((4+c)^2+12)=c^2+8c+28,$ $((4-c)^2+12)=c^2-8c+28.$

Hence $(c^2+28+8c)(c^2+28-8c)=1024,$ $(c^2+28)^2-64c^2=1024.$

So, $c^4-8c^2+784=1024,$ $c^4-8c^2-240=0.$

Let $u=c^2$. Then $u^2-8u-240=0,$ $u=20 \text{ or } -12.$

Since $c^2>0$, $c^2=20.$

Thus, $a^2+b^2=20. \qquad ...(2)$

3. Find $a^2$ and $b^2$

Let $A=a^2,\quad B=b^2.$ Then from (2), $A+B=20. \qquad ...(3)$

From (1), $\frac{16}{A}-\frac{12}{B}=1.$

Using $B=20-A$, $\frac{16}{A}-\frac{12}{20-A}=1.$

Multiply by $A(20-A)$: $16(20-A)-12A=A(20-A).$

So, $320-16A-12A=20A-A^2,$ $320-28A=20A-A^2,$ $A^2-48A+320=0.$

Thus, $A=40 \text{ or } 8.$

Since $A+B=20$, $A=40$ is impossible. Hence $a^2=8,$ $b^2=12.$

4. Compute $p$ and $q$

• Length of conjugate axis of the hyperbola is $p=2b=2\sqrt{12}.$ So, $p^2=4b^2=4\cdot 12=48.$

• Length of latus rectum of the hyperbola is $q=\frac{2b^2}{a}=\frac{2\cdot 12}{\sqrt8}.$ But we need only $q^2$: $q^2=\left(\frac{2b^2}{a}\right)^2=\frac{4b^4}{a^2}.$ Now, $b^4=12^2=144, \quad a^2=8,$ so $q^2=\frac{4\cdot 144}{8}=72.$

Therefore, $p^2+q^2=48+72=120.$

5. Final answer

$\boxed{120}$

This matches the stored correct answer.