1. Find the foci of the given ellipse

The ellipse is $\frac{x^2}{9}+\frac{y^2}{25}=1.$

Since the larger denominator is under $y^2$, its major axis is along the $y$-axis. Thus, $a^2=25,\quad b^2=9.$ So, $a=5,\quad b=3.$ For an ellipse, $c^2=a^2-b^2=25-9=16 \implies c=4.$ Hence the foci are $(0,\pm 4).$

The eccentricity of the ellipse is $e_1=\frac{c}{a}=\frac{4}{5}.$

2. Find the eccentricity of the hyperbola

Given that the eccentricity of the hyperbola is $\frac{15}{8}$ times that of the ellipse: $e_2=\frac{15}{8}\cdot \frac{4}{5}=\frac{3}{2}.$

So the hyperbola has eccentricity $e=\frac{3}{2}.$

3. Determine the hyperbola

The hyperbola has the same foci $(0,\pm 4)$, so its transverse axis must also be along the $y$-axis. Hence its equation is of the form $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1,$ with $c^2=a^2+b^2=16$ and $e=\frac{c}{a}=\frac{3}{2}.$

Since $c=4$, $\frac{4}{a}=\frac{3}{2} \implies a=\frac{8}{3}.$ Thus, $a^2=\frac{64}{9}.$ Now, $b^2=c^2-a^2=16-\frac{64}{9}=\frac{80}{9}.$

So the hyperbola is $\frac{y^2}{64/9}-\frac{x^2}{80/9}=1.$

4. Use focal distance property of hyperbola

For a point on a hyperbola, the absolute difference of distances from the two foci is constant and equal to $2a$. Thus, $|PF_1-PF_2|=2a=\frac{16}{3}.$

We are asked for the smaller focal distance of the point $P\left(\sqrt2,\frac{14}{3}\sqrt{\frac25}\right).$

Let the distances from the foci $(0,4)$ and $(0,-4)$ be $d_1$ and $d_2$. First compute one of them conveniently.

Since $P\left(\sqrt2,\frac{14}{3}\sqrt{\frac25}\right),$ we calculate distance from $(0,4)$: $d_1=\sqrt{(\sqrt2)^2+\left(\frac{14}{3}\sqrt{\frac25}-4\right)^2}.$ That is, $d_1=\sqrt{2+\left(\frac{14}{3}\sqrt{\frac25}-4\right)^2}.$

Now, $\left(\frac{14}{3}\sqrt{\frac25}\right)^2=\frac{196}{9}\cdot\frac25=\frac{392}{45}.$ So,

Hence,

Combine constants: $2+16+\frac{392}{45}=18+\frac{392}{45}=\frac{810+392}{45}=\frac{1202}{45}.$ So, $d_1^2=\frac{1202}{45}-\frac{112}{3}\sqrt{\frac25}.$

Now test option C: $7\sqrt{\frac25}-\frac83.$ Its square is