In the expansion of $(1+px{)}^n,n\in \mathbf{N}\mathbf{,}$ the coefficient of $x$ and $x^2$ are $8$ and $24$ respectively, then

We have 
$\begin{array}{r}(1+px{)}^n=1{+}^n{C}_1\left(px\right){+}^n{C}_2(px{)}^2+\dots \\ =1+npx+\frac{1}{2}n(n-1)p^2{x}^2+\dots \end{array}$
According to the hypothesis,
               $np=8$ and $\frac{1}{2}n(n-1)p^2=24$
Putting $p=8/n$ in the second expression, we get

$\begin{array}{l}\frac{1}{2}n(n-1){\left(\frac{8}{n}\right)}^2=24\Rightarrow \frac{n-1}{n}=\frac{24\times 2}{8\times 8}=\frac{3}{4}\\ \Rightarrow 4n-4=3n\Rightarrow n=4\end{array}$
Putting this value in $np=8$, we get $p=2.$