We know that the number of elements of order $k$ in a finite group $G$ is equal to $\sum cl_{G}(x_{i})$=$\sumG/C_{G}(x_{i})$ such that $x_{i}=k$. It is clear that for a prime $p$ if $p\mid cl_{G}(x_{i})$, then $G$ has an element of order $p$. Now let $p\mid \sum cl_{G}(x_{i})$. My question is: Is $G$ has an element of order $p$?($G$ is not $p$group)

As noted, it is unclear what your sums are over. If the $x_i$ are conjugacy class representatives for all conjugacy classes, then $\sum\mathrm{cl}_G(x_i) = G$, so the question has an affirmative answer by Cauchy's Theorem. More likely is that the $x_i$ are representatives from the conjugacy classes of elements of order $k$; in that case, the answer is "no". Although the example I give in the comments is a $p$group, it is easy to turn it into an example which is not a $p$group: take $p=3$, $k=2$, and let $G=C_2\times C_2\times A$, where $A$ is any nontrivial group of order not divisible by $6$ (e.g., $A=C_5$); then $G$ is not a $p$group, but an element $(a,b,c)$ has order $2$ if and only if $c=1$ and $a$ or $b$ are nontrivial; so $G$ has exactly three elements of order $2$, each of which is its own conjugacy class, so the sum equals $3$. However, $G$ is not divisible by $3$, so $G$ does not have any elements of order $3$. 

