Locus of Midpoint of Normal Segment on Ellipse

The equation of the given ellipse is:
$x^2 + 4y^2 = 16 \implies \frac{x^2}{16} + \frac{y^2}{4} = 1$
Here, $a^2 = 16 \implies a = 4$ and $b^2 = 4 \implies b = 2$.

The eccentricity $e$ of the ellipse is given by:
$e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4}{16}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$

The foci are located at $(\pm ae, 0) = (\pm 2\sqrt{3}, 0)$.
Therefore, the equations of the latus rectums are:
$x = \pm 2\sqrt{3}$

Let the coordinates of any point $P$ on the ellipse in parametric form be:
$P = (4\cos\theta, 2\sin\theta)$

The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $(a\cos\theta, b\sin\theta)$ is:
$\frac{ax}{\cos\theta} - \frac{by}{\sin\theta} = a^2 - b^2$

Substituting $a = 4$ and $b = 2$:
$\frac{4x}{\cos\theta} - \frac{2y}{\sin\theta} = 16 - 4 = 12$

The normal meets the $x$-axis at point $Q$. Setting $y = 0$:
$\frac{4x}{\cos\theta} = 12 \implies x = 3\cos\theta$
Thus, the point $Q$ is $(3\cos\theta, 0)$.

Since $M(h, k)$ is the midpoint of the line segment $PQ$:
$h = \frac{4\cos\theta + 3\cos\theta}{2} = \frac{7}{2}\cos\theta \implies \cos\theta = \frac{2h}{7}$
$k = \frac{2\sin\theta + 0}{2} = \sin\theta \implies \sin\theta = k$

Using the identity $\cos^2\theta + \sin^2\theta = 1$, we obtain:
$\left(\frac{2h}{7}\right)^2 + k^2 = 1$

Thus, the locus of $M(x, y)$ is:
$\frac{4x^2}{49} + y^2 = 1$

To find the intersection of this locus with the latus rectums, substitute $x = \pm 2\sqrt{3}$ into the locus equation:
$\frac{4(\pm 2\sqrt{3})^2}{49} + y^2 = 1$
$\frac{4(12)}{49} + y^2 = 1$
$\frac{48}{49} + y^2 = 1 \implies y^2 = 1 - \frac{48}{49} = \frac{1}{49}$
$y = \pm \frac{1}{7}$

Hence, the locus of $M$ intersects the latus rectums at the points:
$\left(\pm 2\sqrt{3}, \pm \frac{1}{7}\right)$

Therefore, the correct option is C.