Given,

${\left(t^2{x}^{\frac{1}{5}}+\frac{{\left(1-x\right)}^{\frac{1}{10}}}{t}\right)}^{15}$

Now $r^{th}$ term is given by,

$T_{r+1}={}^{15}C_r{\left(t^2{x}^{\frac{1}{5}}\right)}^{15-r}·\frac{{\left(1-x\right)}^{\frac{r}{10}}}{t^r}$

For independent of $t$,

$2\left(15-r\right)-r=0$

$30-2r-r=0$

$\Rightarrow r=10$

So, Maximum value of ${}^{15}C_{10}x\left(1-x\right)$ will be at $x=\frac{1}{2}$$\left(\text{as }\frac{dy}{dx}\left(x-x^2\right)=1-2x \& 1-2x=0\Rightarrow x=\frac{1}{2}\right)$

$={}^{15}C_{10}\frac{1}{2}\times \left(\frac{1}{2}\right)=\frac{3003}{4}$

So, $K=\frac{3003}{4}$ and $8K=6006$