The pumping lemma for regular languages is only a tool for showing that a language is not regular; it cannot be used to show that a language is regular. The two most straightforward ways to demonstrate that a language is regular are (1) to write a regular grammar that generates it, and (2) to design a finite state automaton that recognizes it.
In this case, though, it pays to begin by taking a close look at just what words are in $B$. Consider a word $11x$, where $x\in\{0,1\}^*$: since we can set $k=1$, such a word is automatically in $B$, since $1x$ certainly contains at least one $1$. A word that begins with $1$ is not in $B$ if and only if it has no other $1$’s; and a word that begins with $0$ is not in $B$. Thus,
$B=\{1x\in\{0,1\}^*:x\in\{0,1\}^*\text{ and }x\text{ has at least one }1\}\;,$
the language corresponding to the regular expression $10^*1(0\lor 1)^*$. It’s not at all hard to design a regular grammar that generates this language, or a finite state machine that recognizes it.