1. Equation of a variable line through $P(\sqrt{5},\sqrt{5})$

Let the line through $P$ have slope $m$: $y-\sqrt{5}=m(x-\sqrt{5}).$

A point on this line can be written as $(x,y)=(\sqrt{5},\sqrt{5})+t(1,m),$ where $t$ is a parameter.

If this line meets the ellipse $\frac{x^2}{36}+\frac{y^2}{25}=1$ at points $A,B$, then the corresponding parameter values $t_1,t_2$ satisfy the ellipse equation.

2. Substitute the parametric form into the ellipse

Put $x=\sqrt{5}+t,\qquad y=\sqrt{5}+mt.$ Then $\frac{(\sqrt{5}+t)^2}{36}+\frac{(\sqrt{5}+mt)^2}{25}=1.$

Expanding: $\frac{5+2\sqrt{5}t+t^2}{36}+\frac{5+2m\sqrt{5}t+m^2t^2}{25}=1.$

Multiply by $900$: $25(5+2\sqrt{5}t+t^2)+36(5+2m\sqrt{5}t+m^2t^2)=900.$

So, $125+50\sqrt{5}t+25t^2+180+72m\sqrt{5}t+36m^2t^2=900,$ $(25+36m^2)t^2+\sqrt{5}(50+72m)t-595=0.$

Thus $t_1,t_2$ are roots of $(25+36m^2)t^2+\sqrt{5}(50+72m)t-595=0.$

3. Relate $PA,PB$ to $t_1,t_2$

Since direction vector is $(1,m)$, distance from $P$ to the point corresponding to parameter $t$ is $\sqrt{t^2+(mt)^2}=|t|\sqrt{1+m^2}.$

Hence $PA=|t_1|\sqrt{1+m^2},\qquad PB=|t_2|\sqrt{1+m^2}.$ So $PA\cdot PB=|t_1t_2|(1+m^2).$

From the quadratic, $t_1t_2=\frac{-595}{25+36m^2}.$ Therefore $PA\cdot PB=\frac{595(1+m^2)}{25+36m^2}.$

4. Maximize $PA\cdot PB$

Let $f(m)=\frac{595(1+m^2)}{25+36m^2}.$ Since $595>0$, maximize $g(m)=\frac{1+m^2}{25+36m^2}.$

Write $u=m^2\ge 0$: $g(u)=\frac{1+u}{25+36u}.$ Then g'(u)=\frac{(25+36u)-36(1+u)}{(25+36u)^2}=rac{-11}{(25+36u)^2}<0. So $g(u)$ decreases for $u\ge 0$.

Hence maximum occurs at $u=0\Rightarrow m=0.$

So the required line is horizontal: $y=\sqrt{5}.$

5. Find intersection points with the ellipse

Put $y=\sqrt{5}$ in $\frac{x^2}{36}+\frac{y^2}{25}=1.$ Then $\frac{x^2}{36}+\frac{5}{25}=1,$ $\frac{x^2}{36}+\frac15=1,$ $\frac{x^2}{36}=\frac45,$ $x^2=\frac{144}{5}.$ So $x=\pm \frac{12}{\sqrt{5}}.$

Thus $A\left(\frac{12}{\sqrt{5}},\sqrt{5}\right),\qquad B\left(-\frac{12}{\sqrt{5}},\sqrt{5}\right).$

Also $P=(\sqrt{5},\sqrt{5}).$ Since all three points have same $y$-coordinate, distances are horizontal: $PA=\frac{12}{\sqrt{5}}-\sqrt{5}=\frac{12-5}{\sqrt{5}}=\frac7{\sqrt{5}},$ $PB=\sqrt{5}-\left(-\frac{12}{\sqrt{5}}\right)=\frac{5+12}{\sqrt{5}}=\frac{17}{\sqrt{5}}.$

Therefore $PA^2+PB^2=\frac{49}{5}+\frac{289}{5}=\frac{338}{5}.$ Hence $5(PA^2+PB^2)=338.$

6. Check with options

The value is $\boxed{338}$ which is Option C.

7. Comparison with stored answer

Stored correct answer: C

Derived answer: C

So they agree.