Which of the statements is/are TRUE? 

Let $K = \big [k_{i,j} \big ]_{i,j = 1}^{\infty}$ be an infinite matrix over $\Bbb C$ (the set of all complex numbers) such that 
$(\text {i})$ for each $i \in \Bbb N$ (the set of all natural numbers), the $i^{\text {th}}$ row $(k_{i,1},k_{i,2}, \cdots )$ of $K$ is in $\ell^{\infty}$ and 
$(\text {ii})$ for every $x = (x_1,x_2, \cdots) \in \ell^{1},$ $\sum\limits_{j = 1}^{\infty} k_{i,j}\ x_j$ is summable for all $i \in \Bbb N,$ and $(y_1,y_2, \cdots ) \in \ell^{1}$ where $y_i = \sum\limits_{j = 1}^{\infty} k_{i,j}\ x_j.$
Let the set of all rows of $K$ be denoted by $E.$ Consider the following statements $:$
$\text {P}\ :\ $ $E$ is a bounded set in $\ell^{\infty}.$
$\text {Q}\ :\ $ $E$ is a dense set in $\ell^{\infty}.$
$$\begin{align*} \left (\ell^{1}  = \left \{ \left (x_1,x_2, \cdots \right )\ :\ x_i \in \Bbb C, \sum\limits_{i=1}^{\infty} \left |x_i \right | < \infty \right \} \right ) \\ \left ( \ell^{\infty}  = \left \{ \left (x_1,x_2, \cdots \right )\ :\ x_i \in \Bbb C, \sup\limits_{i \in \Bbb N} \left |x_i \right | < \infty \right \} \right   ) \end{align*}$$
Which of the above statements is/are TRUE $?$
$(\text A)$ Both $\text P$ and $\text Q.$
$(\text B)$ $\text P$ only.
$(\text C)$ $\text Q$ only.
$(\text D)$ Neither $\text P$ nor $\text Q.$


How do I solve this problem? Any help will be highly appreciated.
Thank you very much.
EDIT $:$ What I can see is that the column vectors of $K$ are in $\ell^{1}.$ Can it help anyway? I think $E$ is bounded but for that we need only to show that the sequence $\bigg \{ \sup\limits_{j \in \Bbb N} \left |k_{i,j} \right | \bigg \}_{i=1}^{\infty}$ of real numbers is bounded. How do I show that?
Suppose the sequence $\bigg \{ \sup\limits_{j \in \Bbb N} \left |k_{i,j} \right | \bigg \}_{i=1}^{\infty}$ of real numbers is unbounded then we can find a strictly increasing sequence $\big \{i_r \big \}_{r=1}^{\infty}$ of natural numbers such that $$\sup\limits_{j \in \Bbb N} \left |k_{i_r,j} \right | > r,\ \text {for each}\ r \in \Bbb N.$$ So $\exists$ another sequence $\big \{j_r \big \}_{r = 1}^{\infty}$ of natural numbers such that $$\left |k_{i_r,j_r} \right | > r,\ \text {for each}\ r \in \Bbb N.$$ Now this would yield a sequence $\bigg \{\left |k_{i_r,j_r} \right | \bigg \}_{r = 1}^{\infty}$ of real numbers such that $$\left |k_{i_r,j_r} \right | > r,\ \text {for each}\ r \in \Bbb N.\ \ \ \ \ \ (1)$$
Now how do I proceed? I think that I have to somehow use condition $(\text {ii} )$ here. Since the column vectors of $K$ are in $\ell^1$ we have for any fixed $j \in \Bbb N$ $$\lim\limits_{r \to \infty} \left |k_{i_r,j} \right | = 0.$$
So in particular if we take $j=j_r$ then we have $$\lim\limits_{r \to \infty} \left | k_{i_r,j_r} \right | = 0$$ which contradicts $(1).$ Is it correct what I have argued? Please verify it. 
I don't think the last argument is true. The matrix $K$ may be such that for each $r \in \Bbb N,$ $j_r$'s could have been distinct and $k_{i_r,j_r}$ is the last non-zero entry in the $j_r^{\text {th}}$ column and $\left |k_{i_r,j_r} \right | = r + 1 > r.$ That's why I get stuck and can't able to proceed further.
 A: $E$ is countable and thus never dense in the non-separable space $\ell^\infty$. 
Asumption (ii) says that $T:\ell^1\to \ell^1$, $T(x)=\left(\sum\limits_{j=1}^\infty k_{i,j}x_j\right)_{i\in\mathbb N}$ is a well-defined map which is obviously linear. Show that its graph is closed (because it is continuous with respect to the weaker topology of coordinate-wise convergence on the range) to conclude its continuity. This gives you a uniform bound 
$$
c=\sup\{\sum_{i=1}^\infty\left|\sum_{j=1}^\infty k_{i,j}x_j\right|: x\in\ell^1,\|x\|_1\le 1\}<\infty.
$$
For fixed $i_0,j_0$ plug in the sequence $x$ which has only $0$ entries but at the $j_0$-th spot where the entry is the sign of $k_{i_0,j_0}$ (in the complex case, the sign of $z\in\mathbb C\setminus\{0\}$ is $|z|/z$). This gives $|k_{i_0,j_0}|\le c$ for all $(i_0,j_0)$ and hence the boundedness of the set of rows in $\ell^\infty$.

Instead of the closed graph theorem it is also possible to use the uniform boundedness principle for the maps $T_i(x)=\sum\limits_{j=1}^\infty k_{i,j}x_j$. This would require an assumption  weaker than (ii).
