Let the infinite GP be $a, ar, ar^2, ar^3, \dots$ with first term $a$ and common ratio $r$, where $|r|<1$.

Given sum of the GP: $\frac{a}{1-r}=15 \quad \Rightarrow \quad a=15(1-r).$

The squares of the terms form another GP: $a^2, a^2r^2, a^2r^4, \dots$ whose sum is $\frac{a^2}{1-r^2}=150.$

Substitute $a=15(1-r)$ into this equation: $\frac{[15(1-r)]^2}{1-r^2}=150.$ $\frac{225(1-r)^2}{(1-r)(1+r)}=150.$ $\frac{225(1-r)}{1+r}=150.$ Divide by $75$: $3\frac{1-r}{1+r}=2.$ $\frac{1-r}{1+r}=\frac{2}{3}.$

Solve for $r$: $3(1-r)=2(1+r)$ $3-3r=2+2r$ $1=5r$ $r=\frac{1}{5}.$

Then $a=15\left(1-\frac15\right)=15\cdot \frac45=12.$

We need the sum of $ar^2+ar^4+ar^6+\dots$ This is a GP with first term $ar^2$ and common ratio $r^2$.

So, $S=\frac{ar^2}{1-r^2}.$

Substitute $a=12$ and $r=\frac15$: $S=\frac{12\cdot \left(\frac{1}{25}\right)}{1-\frac{1}{25}}$ $=\frac{12/25}{24/25}$ $=\frac{12}{24}=\frac12.$

Therefore, the required sum is $\boxed{\frac12}.$

Comparing with the given options, this is Option B.