It's not an Analytic Geometry problem

Let's look at the ellipse: $\frac { { x }^{ 2 } }{ 4 } +\frac { { y }^{ 2 } }{ 3 } =1$ . It's semi-axis's are $a=2,b=\sqrt { 3 }$ , which means the focal distance is $c=\sqrt { { a }^{ 2 }-{ b }^{ 2 } } =\sqrt { 4-3 } =1$ . So, the point $F(-1,0)$ is either a focal point of the ellipse and a center of the circle:

The circle is fully inside the ellipse because the closest point of the ellipse to point F is $a-c=1$ far. That means, $\left| \overrightarrow { PF } \right|$ varies from $1$ to $3$ .

Now let's consider the tangent lines, $AP$ and $BP$ :

Since they are tangent from the same point, they are perpendicular to the radii, $AF$ and $BF$ , and $AP=BP$ . That means triangles $\triangle APF$ and $\triangle BPF$ are equal. So, $\angle APB = \angle APF + \angle FPB = 2\angle APF$ .

Let $PF = d$ and $\angle APF=\alpha$ , then:

$\left| \overrightarrow { PF } \right| ^{ 2 }=\left| \overrightarrow { AF } \right| ^{ 2 }+\left| \overrightarrow { PA } \right| ^{ 2 }\\ { d }^{ 2 }=1+{ \left| \overrightarrow { PA } \right| }^{ 2 }\\ \left| \overrightarrow { PA } \right| =\sqrt { 1-{ d }^{ 2 } }$

$\cos { \alpha } =\frac { \left| \overrightarrow { PA } \right| }{ \left| \overrightarrow { PF } \right| } =\frac { \sqrt { { d }^{ 2 }-1 } }{ d } \\ \cos { \alpha } =\sqrt { 1-\frac { 1 }{ { d }^{ 2 } } }$

$\cos { \left( 2\alpha \right) } =2\cos ^{ 2 }{ \alpha } -1\\ \cos { \left( 2\alpha \right) } =1-\frac { 2 }{ { d }^{ 2 } }$

Let's derive the needed value:

$\overrightarrow { PA } \cdot \overrightarrow { PB } =\cos { \angle APB } \left| \overrightarrow { PA } \right| \left| \overrightarrow { PB } \right| \\ \overrightarrow { PA } \cdot \overrightarrow { PB } =\cos { \left( 2\alpha \right) } { \left| \overrightarrow { PA } \right| }^{ 2 }\\ \overrightarrow { PA } \cdot \overrightarrow { PB } =\left( 1-\frac { 2 }{ { d }^{ 2 } } \right) \left( 1-{ d }^{ 2 } \right) \\ \overrightarrow { PA } \cdot \overrightarrow { PB } ={ d }^{ 2 }-3+\frac { 2 }{ { d }^{ 2 } }$

Now we can find the range, knowing $d\in \left[ 1,3 \right]$ . Let's find the derivative:

$\frac { \partial \left( \overrightarrow { PA } \cdot \overrightarrow { PB } \right) }{ \partial d } =2d-\frac { 4 }{ { d }^{ 3 } }$

$2d-\frac { 4 }{ { d }^{ 3 } } =0\\ 2{ d }^{ 4 }=4\\ d={ 2 }^{ \frac { 1 }{ 4 } }$

So we've got:

$\begin{cases} \frac { \partial \left( \overrightarrow { PA } \cdot \overrightarrow { PB } \right) }{ \partial d } <0,\quad 1\le d<{ 2 }^{ \frac { 1 }{ 4 } } \\ \frac { \partial \left( \overrightarrow { PA } \cdot \overrightarrow { PB } \right) }{ \partial d } =0,\quad d={ 2 }^{ \frac { 1 }{ 4 } } \\ \frac { \partial \left( \overrightarrow { PA } \cdot \overrightarrow { PB } \right) }{ \partial d } >0,\quad { 2 }^{ \frac { 1 }{ 4 } }<d\le 3 \end{cases}$

That means minimum is reached at $d={ 2 }^{ \frac { 1 }{ 4 } }$ and the maximum is at one of the ends of the interval, $d=1$ or $d=3$ :

${ \overrightarrow { PA } \cdot \overrightarrow { PB } }_{ d={ 2 }^{ \frac { 1 }{ 4 } } }=\sqrt { 2 } -3+\sqrt { 2 } =2\sqrt { 2 } -3\\ { \overrightarrow { PA } \cdot \overrightarrow { PB } }_{ d=1 }=1-3+2=0\\ { \overrightarrow { PA } \cdot \overrightarrow { PB } }_{ d=3 }=9-3+\frac { 2 }{ 9 } =\frac { 56 }{ 9 }$

So, $l=2\sqrt { 2 } -3$ and $r=\frac { 56 }{ 9 }$ and the final answer is $\boxed { 60506 }$