First, for any Banach Space $X$ and a weakly convergent sequence $x_{n}$, define $T_{n}\in X^{**}$ by $T_{n}(f)=f(x_{n})$.
Then see that $T_{n}$'s are linear and continuous. Now, for each $f$, $\sup_{n}|T_{n}(f)|<C_{f}$ for some $C_{f}$ (depending on $f$) as $f(x_{n})$ is convergent. Thus the Uniform Boundedness Principle tells you that $\sup_{n}||T_{n}||<\infty$. But $||T_{n}||=\sup_{||f||\leq 1}|f(x_{n})|$
Thus $\sup_{n}\sup_{||f||\leq 1}|f(x_{n})|<\infty$ . You can apply the Hahn-Banach Theorem to show that $\sup_{||f||\leq 1}|f(x)|=||x||$ for each $x\in X$.
i.e. you always have $|f(x)|\leq ||x||$ if $||f||=1$. But you can define $g:(\Bbb{K}x)\to\Bbb{K}$ by $g(tx)=t||x||$. Then $||g||\leq 1$ and $g$ is linear and thus, by Hahn-Banach Theorem, there exists $G:X\to\Bbb{K}$ such that $||G||\leq 1$ and $G|_{(\Bbb{K}x)}=g$ . Thus you have that $G(x)=||x||$ and $\sup_{||f||\leq 1}|f(x)|=||x||$.
This shows you that $\sup_{n}\sup_{||f||\leq 1}|f(x_{n})|=\sup_{n}||x_{n}||<\infty$ .
Thus you have $(a)$. For $(b)$ you only need to note that $x^{n}_{i}=\langle e_{i},x^{n}\rangle\to \langle e_{i},x\rangle=x_{i}$ because $e_{i}\in (\ell^{p})^{*}$.
Converse
Note that if $\mu$ denotes the counting measure on $\Bbb{N}$, then $\displaystyle\int_{\Bbb{N}} |x|^{p}\,d\mu=\sum_{j=1}^{\infty}|x_{j}|^{p}$ for any $x\in\ell^{p},p<\infty$
Let $(a)$ and $(b)$ hold. And let $p=\infty$ for the while. We'll show that for $y\in\ell^{1}$, you have $\sum_{j=1}^{\infty}y_{j}x^{n}_{j}\xrightarrow{n\to\infty}\sum_{j=1}^{\infty}y_{j}x_{j}$ which is what is asked.
Note : This is not the same as saying that for all $f\in(\ell^{\infty})^{*}$, you have that $\langle f,x^{n}\rangle\to \langle f,x\rangle$ . That is false. But, for the particular topology $\sigma(\ell^{\infty},\ell^{1})$ ( weak$^*$ topology in the dual of $\ell^{1}$), you do have what is asked in the question. The point being that the weak$^{*}$ topology on $\ell^{\infty}$ is not the same as
the weak topology on $\ell^{\infty}$.
Note that you have $\sup_{n}\sup_{j}|x_{j}^{n}|<\infty$ and thus $|x_{j}^{n}|\leq C$ for all $j$ and $n$ for some constant $C$.
So, you have that $|y_{j}x^{n}_{j}|\leq C|y_{j}|$ and $\displaystyle\int_{\Bbb{N}}|y|\,d\mu=\sum_{j=1}^{\infty}|y_{j}|<\infty$ .
Thus, as $y_{j}x^{n}_{j}\to y_{j}x_{j}$, By the Dominated Convergence Theorem, you have that
$$\int_{\Bbb{N}}yx^{n}\,d\mu=\sum_{j=1}^{\infty}y_{j}x^{n}_{j}\xrightarrow{n\to\infty}\sum_{j=1}^{\infty}y_{j}x_{j}=\int_{\Bbb{N}}yx\,d\mu$$
Thus $x^{n}\xrightarrow{w^{*}}x$ for $p=\infty$
For $p=1$, unfortunately, the conclusion is false and it seems to be a mistake in the book.
For a simple counter example, let $x^{n}$ be $e_{n}$. Then $x^{n}_{i}$ converges to $0$ for each $i$ and $||x^{n}||=1$ for each $n$. But for the functional $1\in\ell^{\infty}$, you have $\langle 1,x^{n}\rangle=\sum_{j=0}^{\infty}x^{n}_{j}=1$ which is not $\langle 1,0\rangle=\sum_{j=1}^{\infty} 0\cdot 1=0$
Infact, strong convergence is the same as weak convergence in $\ell^{1}$ due to a Theorem of Schur so you don't have strong or weak convergence in this case.
Now let $1<p<\infty$. Now $\ell^{p}$ is Reflexive.
First note that by Fatou's Lemma and assumption (a) $$\int_{\Bbb{N}}|x|^{p}\,d\mu=\sum_{j=1}^{\infty}|x_{j}|^{p}\leq \lim_{n\to\infty}\inf\sum_{j=1}^{\infty}|x^{n}_{j}|^{p}<\infty$$ as $\sup_{n}||x^{n}||_{\ell^{p}}<\infty$. Thus $x\in\ell^{p}$
Now use Theorem $3.18$ in Brezis which states that every bounded sequence in a reflexive Banach space has a weakly convergent subsequence.
So, let $x^{n_{k}}$ be an arbitrary subsequence of $x^{n}$. Then, you have that $\sup_{k}||x^{n_{k}}||_{\ell^{p}}$ is bounded and thus, there exists a further subsequence $x^{n_{k_{l}}}$ which converges weakly to some $z$ (say).
And hence $\langle e_{i},x^{n_{k_{l}}}\rangle\to z_{i}$. But, you already have that $$\langle e_{i},x^{n_{k_{l}}}\rangle=x^{n_{k_{l}}}_{i}\to x_{i}$$
Thus $x_{i}=z_{i}$ for all $i$ and hence $x=z$.
So, what we have now shown is that given any subsequence of $x^{n}$, there exists a further subsequence which converges weakly to $x$. I claim that this is sufficient to show that $x^{n}\rightharpoonup x$. This would completes the exercise. (This is a very useful criteria for checking weak convergence). Try to prove it. If you can't then see below
So let $X$ be any Normed Space and $f\in X^{*}$ and let $x_{n}$ be a
sequence such that given any subsequence, there exists a further
subsequence which converges weakly to $x$.
Then you have that the real/complex sequence $f(x_{n})$ satisfies the
property that any given subequence of it has a further subequence
which converges to $f(x)$. This is a critria for sequences in a
metric space to converge.
Thus $f(x_{n})\to f(x)$ and this holds for all $f$ and hence
$x_{n}\rightharpoonup x$
Alternatively, As @geetha290krm suggests,
Let $y\in\ell^{p'}$.
Now, by Holder's Inequality, and using that $\sup_{n}||x^{n}||_{\ell^{p}}<\infty$ $$\sum_{k=m}^{\infty}|y_{k}(x^{n}_{k}-x_{k})|\leq (\sum_{k=m}^{\infty}|y_{k}|^{p'})^{1/p'}(\sum_{k=m}^{\infty}|x^{n}_{k}-x_{k}|^{p})^{1/p}\leq C(\sum_{k=m}^{\infty}|y_{k}|^{p'})^{1/p'}$$
where $C$ is a constant.
Thus, given $\epsilon>0$, you can find an $M$(depending on $\epsilon$) such that $m\geq M$ implies $\sum_{k=m}^{\infty}|y_{k}(x^{n}_{k}-x_{k})|<\epsilon$. This is because $\sum_{k=m}^{\infty}|y_{k}|^{p'}\to 0$ as $m\to\infty$.
Now $\displaystyle\bigg|\sum_{k=1}^{\infty}y_{k}(x_{k}^{n}-x_{k})\bigg|=\bigg|\sum_{k=1}^{M}y_{k}(x_{k}^{n}-x_{k})+\sum_{k=M}^{\infty}y_{k}(x_{k}^{n}-x_{k})\bigg|$
Now choose $N$ (depending on $\epsilon$ and $M$ and hence only on $\epsilon$) such that $n\geq N$ implies $\bigg|\sum_{k=1}^{M}y_{k}(x_{k}^{n}-x_{k})\bigg|<\epsilon$ (this can be done due to coordinate wise convergence).
Thus, for all $n\geq N$, you have that
$$\bigg|\sum_{k=1}^{\infty}y_{k}(x_{k}^{n}-x_{k})\bigg|<2\epsilon$$ which shows that $\langle y,x^{n}\rangle\to \langle y,x\rangle$
