Prove that $\delta_K(x) \in Z(K(G))$ Let $G$ be finite group and $K$ will be conjugacy class of $G$.
And $\delta_K(x) = \begin{equation*} 
 \begin{cases}
   1, x \in K\\
   0, x \notin K
 \end{cases}
\end{equation*}$
How to prove that $\delta_K \in Z(K(G))$ and find basis of $K(G)$, where $K(G)$ is group algebra over a finite group. And $Z(K(G))$ is center of a group algebra
 A: You are mixing some notations above. Nevertheless, let $C$ be a conjugacy class in $G$ then your $\delta_C$ is $\sum_{g\in C}g$. It obviously in the center of $kG$, since $G$ acts on $C$ by permutations and hence for some $h\in G$
$$\sum_{g\in C}g = \sum_{hgh^{-1}\in C}hgh^{-1} = \sum_{g\in C}hgh^{-1} =  h(\sum_{g\in C}g)h^{-1}.$$
The basis of $Z(kG)$ is given by the elements $\delta_{C_i} = \sum_{g\in C_i}g$, where $C_1,C_2, \dots , C_n$ are conjugacy classes of $G$. It is obvious from above that ${\rm span} (\delta_{C_1}, \delta_{C_2}, \dots, \delta_{C_n}) \subset Z(kG).$ Conversely let $a\in Z(kG).$ Then $a = \sum_{g\in G} a_gg$ and $hah^{-1} = a$ for $h\in G$, i.e. 
$$\sum_{g\in G}a_g g = \sum_{g\in G}a_g hgh^{-1}.$$ Since $\{hgh^{-1} \mid  g\in G\} = G$ we have for $f=hgh^{-1}$ and $g=h^{-1}fh$ $$\sum_{g\in G}a_g g = \sum_{g\in G}a_g hgh^{-1}=\sum_{f\in G}a_{h^{-1}fh} f.$$ This gives $a_g = a_{h^{-1}gh}$, i.e. $a_g$ is constant every conjugacy class. Hence for representatives $g_i \in C_i$
$$a = \sum_{i=1}^na_{g_i}\delta_{C_i}.$$
This proves ${\rm span}(\delta_{C_1}, \delta_{C_2}, \dots, \delta_{C_n})\subset Z(kG).$
