It's not that difficult!

Now, $\displaystyle \sum _{ r=0 }^{ n }({\displaystyle \sum _{ p=0 }^{ r-1 }{ {n \choose r}{r \choose p} 2^{p} }) } = \displaystyle \sum _{ r=0 }^{ n }({n \choose r}{\displaystyle \sum _{ p=0 }^{ r-1 }{ {r \choose p} 2^{p} } ) }$

But $\displaystyle \sum _{ p=0 }^{ r }{ {r \choose p} 2^{p} } = (1+2)^{r}$

Thus, $\displaystyle \sum _{ p=0 }^{ r-1 }{ {r \choose p} 2^{p} } = (1+2)^{r}-2^{r}$

Hence we have $\displaystyle \sum _{ r=0 }^{ n }({n \choose r}{\displaystyle \sum _{ p=0 }^{ r-1 }{ {r \choose p} 2^{p} } ) }$ $=$ $\displaystyle \sum _{ r=0 }^{ n }{ {n \choose r}(3^{r}-2^{r})}$

$= \displaystyle \sum _{ r=0 }^{ n }{ {n \choose r}3^{r}}- \displaystyle \sum _{ r=0 }^{ n }{ {n \choose r}2^{r}}$

$= (1+3)^{n} - (1+2)^{n} = \boxed{4^{n}-3^{n}}$

We can 'cheat' using the multiple choice options; for $n=1$ the double sum contains only one term which is equal to $1$ , therefore $\boxed{4^n-3^n}$ is the only possibility (since $4^1-3^1=1$ ).