Let $\{c_{k}\}_{k\in \mathbb Z} \subset \mathbb C$ such that, $\sum_{k\in \mathbb Z} |c_{k}| < \infty.$ Let $\delta_{k}$ is the unit Dirac mass at $k $, we note that $\mu = \sum_{k\in \mathbb Z} c_{k} \delta_{k}$ is complex Borel measure on $\mathbb R.$
My Question is: Is it true that, $\int_{\mathbb R} e^{-ixy} \sum_{k\in \mathbb Z} c_{k} d\delta_{k}(x) = \sum_{k\in \mathbb Z} c_{k}\int_{\mathbb R} e^{ixy} d\delta_{k}(x);$ for fix $y\in \mathbb R;$ if yes, how to justify it ?
Yes, it is true. Define a (complex-valued) measure on $\mathbb{R}$ by
$$\mu(B) := \sum_{k \in \mathbb{Z}} c_k \delta_k(B), \qquad B \in \mathcal{B}(\mathbb{R}). \tag{1}$$
Since $\sum_{k \in \mathbb{Z}} |c_k|<\infty$, $\mu$ is a finite measure on $\mathbb{R}$. From $(1)$ we see that the formula
$$\int f(x)\left( \sum_{k \in \mathbb{Z}} c_k \delta_k(dx) \right) = \sum_{k \in \mathbb{Z}} c_k \int f(x) \delta_k(dx) \tag{2}$$
holds for any $f=1_B$ where $B \in \mathcal{B}(\mathbb{R})$ is a Borel set. Since the integral is linear, this means that $(2)$ holds for any step function of the form
$$f(x) = \sum_{j=1}^n f_j \cdot 1_{B_j}(x), \qquad f_j \in \mathbb{C},B_j \in \mathcal{B}(\mathbb{R}).$$
For any positive (real-valued) integrable function $f$, there exists a sequence of step functions $(f_n)_n$ such that $f_n(x) \uparrow f(x)$ for any $x \in \mathbb{R}$. Using the monotone convergence theorem, we see that $(2)$ holds also for $f$. Finally, we can split up any (complex-valued) function as $$f= (f^+_r - f^-_r) + \imath (f_i^+ - f_i^-)$$ where $f^+_r$ (resp. $f^-_r$) denotes the positive (resp. negative) part of $\text{Re} \, f$ and $f^+_i$ (resp. $f^-_i$) denotes the positive (resp. negative) part of $\text{Im} \, f$. Therefore, it follows from the linearity of the integral that $(2)$ holds for any (integrable) function $f$.