Given:

$9x^2+16y^2=144$

$\Rightarrow \frac{x^2}{16}+\frac{y^2}{9}=1$

Equation of tangent at point $P\left(4\cos \theta , 3\sin \theta \right)$ is 

$\frac{x\cos \theta }{4}+\frac{y\sin \theta }{3}=1$

So, $A$ is $\left(4\sec \theta ,0\right)$ and point $B$ is $\left(0,3cosec\theta \right)$.

Hence,

$AB=\sqrt{16 {\sec }^2\theta +9{cosec}^2\theta }$

$\Rightarrow AB=\sqrt{25+16{\tan }^2 \theta +9{\cot }^2 \theta }$

$\Rightarrow \sqrt{25+16{\tan }^2 \theta +9{\cot }^2 \theta }\ge 7$