1. Find points $A$ and $B$ on the line $5x+7y=50$

Since $A(\alpha,0)$ lies on $5x+7y=50$: $5\alpha=50 \implies \alpha=10$ So, $A=(10,0)$

Since $B(0,\beta)$ lies on the same line: $7\beta=50 \implies \beta=\frac{50}{7}$ So, $B=\left(0,\frac{50}{7}\right)$

2. Find point $P$ dividing $AB$ internally in the ratio $7:3$

Using section formula: if $P$ divides $AB$ internally in the ratio $7:3$, then $P=\left(\frac{7x_2+3x_1}{7+3},\frac{7y_2+3y_1}{7+3}\right)$ where $A(x_1,y_1)=(10,0), \quad B(x_2,y_2)=\left(0,\frac{50}{7}\right)$

Thus, $x_P=\frac{7\cdot 0+3\cdot 10}{10}=3$ $y_P=\frac{7\cdot \frac{50}{7}+3\cdot 0}{10}=\frac{50}{10}=5$ So, $P=(3,5)$

3. Use the directrix information

The ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ Given directrix: $3x-25=0 \implies x=\frac{25}{3}$

For an ellipse with major axis along the $x$-axis, the directrices are $x=\pm \frac{a}{e}$ where $e$ is eccentricity, and the foci are $S=(\pm ae,0)$

Since the perpendicular from $S$ to the $x$-axis passes through $P$, and the $x$-axis is horizontal, that perpendicular is a vertical line through $S$. Hence the foot of this perpendicular is $(x_S,0)$, and it passes through $P=(3,5)$ only if $x_S=3$ Therefore the corresponding focus is $S=(3,0)$ So, $ae=3$

Also from the directrix, $\frac{a}{e}=\frac{25}{3}$

4. Find $a$ and $e$

Multiply the two equations: $(ae)\left(\frac{a}{e}\right)=3\cdot \frac{25}{3}=25$ $a^2=25 \implies a=5$

Then, $e=\frac{3}{5}$

5. Find $b^2$

For an ellipse, $b^2=a^2(1-e^2)$ So, $b^2=25\left(1-\frac{9}{25}\right)=25\cdot \frac{16}{25}=16$

6. Find the length of the latus rectum

For the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, the length of the latus rectum is $\frac{2b^2}{a}$ Hence, $\text{Length of latus rectum}=\frac{2\cdot 16}{5}=\frac{32}{5}$

7. Final answer

$\boxed{\frac{32}{5}}$ This corresponds to Option C.