Given Ellipse $\frac{x^2}{16}+\frac{y^2}{4}=1$

$e=\sqrt{1-\frac{b^2}{a^2}}=\frac{\sqrt{3}}{2}$

$\because P$ is a point on the ellipse

So, $P =(4 \cos \theta, 2 \sin \theta)$

Equation of normal to the ellipse $\frac{x^2}{16}+\frac{y^2}{4}=1$ at point $\left( x _1, y _1\right)=(4 \cos \theta, 2 \sin \theta)$ is given by

$a ^2 y _1\left( x – x _1\right)= b ^2 x _1\left( y – y _1\right)$

$\Rightarrow 16 \times 2 \sin \theta( x -4 \cos \theta)=4 \times 4 \cos \theta( y -2 \sin \theta)$

$\Rightarrow 2 x \sin \theta-8 \sin \theta \cos \theta= y \cos \theta-2 \sin \theta \cos \theta)$

$\Rightarrow 2 x \sin \theta= y \cos \theta+6 \sin \theta \cos \theta$

$\Rightarrow \frac{2 x }{\cos \theta}=\frac{ y }{\sin \theta}+6$

$\Rightarrow 2 x \sec \theta- y \operatorname{cosec} \theta=6$

It meet the $x$-axis at $Q(3 \cos \theta, 0)$

$\therefore M =\left(\frac{7}{2} \cos \theta, \sin \theta\right)=( x , y )$

Locus of $M$ is

$\frac{x^2}{\left(\frac{7}{2}\right)^2}+\frac{y^2}{1}=1$

Latus rectum of the given ellipse is

$x = \pm ae = \pm \sqrt{16-4}= \pm 2 \sqrt{3}$

So locus of $M$ meets the latus rectum at points for which

$y ^2=1-\frac{12 \times 4}{49}=\frac{1}{49} \Rightarrow y = \pm \frac{1}{7}$

Hence, the required point is $\left( \pm 2 \sqrt{3}, \pm \frac{1}{7}\right)$.