Given $a=(a_i)_{i=1}^\infty$ with $a_i \geq 0$ and $b=(b_i)_{i=1}^\infty$ with $b_i \in \mathbb{R}$, let $$E_i = \lbrace (x_n)_{n=1}^\infty : n^{b_i}|x_n|\leq a_i, \forall n\in \mathbb{N} \rbrace$$ and $E= (E_i)_{i=1}^\infty$
The question ask for criterion for the sequences $a$ and $b$ such that $E$ to be bounded in ${\oplus}_{p=1}^\infty \ell_p$ and precompact(totally bounded) in $\Pi_{p=1}^\infty \ell_p$.
My first thought was that to use Tychonoff's Theorem(The product of any collection of compact topological spaces is compact with respect to the product topology) since compactness require boundedness and precompactness. But couldn't figure it out.
I am studying for the midterm which in one week, so I am grateful for any help/hints.
It seems the following.
Put $\mathcal L={\oplus}_{p=1}^\infty \ell_p$ and $\widehat{\mathcal L}=\Pi_{p=1}^\infty \ell_p$. There can be many norms on $\mathcal L$, but at least I expect that the restriction $\|\cdot\||_{\ell_p}$ of the norm $\|\cdot\|$ which you consider on $\mathcal L$, coincides with the standard norm $\|\cdot\|_p$ of $\ell_p$, which is naturally embedded into $\mathcal L$.
I consider $\widehat{\mathcal L}$ as a topological group endowed with a Tychonoff product topology. It is easy to check that the set $E$ considered as a direct sum ${\oplus}_{p=1}^\infty E_p$ (or, even a product $\Pi_{p=1}^\infty E_p$) is totally bounded in $\widehat{\mathcal L}$ iff each set $E_p$ is totally bounded in $\ell_p$.
The set $E_p$ belongs to $\ell_p$ iff $M_p=\sum_{n=1}^\infty \left(\frac{a_p}{n^{b_p}}\right)^p<\infty$, that is iff $a_p=0$ or $pb_p>1$. If so, then the set $E_p$ is bounded in $\ell_p$ (because $\|x\|_p\le M_p^{1/p}$ for each sequence $x\in E_p$). So in order that $E$ is bounded in $\mathcal L$ we have to have $M=\sup_p M_p^{1/p}<\infty$. If so, then the boundedness of the $E$ in $\mathcal L$ depends on the norm $\|\cdot\|$. For instance, let $y=(y_p)\in\mathcal L$. If $\|y\|=\sup_p \|y_p\|$ for $y=(y_p)\in\mathcal L$, then $\|z\|\le M$ for each $z\in E$. If $\|y\|=\sum_p \|y_p\|$, then $E$ is bounded in $\mathcal L$ iff all but finitely many $M_p$ are zeroes. If $\|y\|=\left(\sum_p \|y_p\|^q\right)^{1/q}$ , then $E$ is bounded in $\mathcal L$ iff $\sum_p M_p^{q/p}<\infty$.
At last, the set $E_p$ is totally bounded in $\ell_p$ iff $M_p=0$. Indeed, if $M_p=0$ then $E_p=\{0\}$ is a totally bounded set. If $M_p>0$ then $a_p>0$, so the set $E_p$ contains a sequence $\{e^{p,m}\}$ (because $0\le a_p$ and $n^{b_p}|a_p|\le n^0|a_p|=a_p$), where $e^{p,m}_n=0$, if $m\ne n$ and $e^{p,m}_n=a_p$, if $m=n$. But if $m\ne m’ $ then $\|e^{p,m}- e^{p,m’}\|=\left(2a_p^p\right)^{1/p}=2^{1/p}a_p$. So the set $E_p$ is not totally bounded.