An easier way is to recognize that $X = Y_1 + Y_2 + \cdots Y_n$ where $Y_k$ are independent Bernoulli random variables with parameter $p$. For a Bernoulli random variable $Y_k$, we have $\text{Var}(Y_k) = p(1-p)$ Since $Y_k$ are independent, we have that $\text{Var}(X) = \text{Var}(Y_1) + \text{Var}(Y_2) + \cdots + \text{Var}(Y_n) = np(1-p)$
To go the direct way, we need to first evaluate couple of summations.
We will evaluate the sums $\sum_{k = 0}^n k \mathbb{P}(X=k) \text{ and }\sum_{k = 0}^n k^2 \mathbb{P}(X=k)$ First note that $\mathbb{P}(X=k) = \dbinom{n}k p^k (1-p)^{n-k}$. Hence, $\sum_{k = 0}^n k \mathbb{P}(X=k) = \sum_{k = 0}^n k \dbinom{n}k p^k (1-p)^{n-k}$ Note that $k \dbinom{n}k = \dfrac{n!}{(n-k)! (k-1)!} = n \dbinom{n-1}{k-1}$ Hence, \begin{align} \sum_{k = 0}^n k \mathbb{P}(X=k) & = \sum_{k = 1}^n n \dbinom{n-1}{k-1} p^k (1-p)^{n-k} = np \sum_{k = 1}^n \dbinom{n-1}{k-1} p^{k-1} (1-p)^{(n-1)-(k-1)}\\ & = np \left( p + (1-p)\right)^{n-1} = np \end{align} Similarly, \begin{align} k^2 \dbinom{n}k & = k \dfrac{n!}{(n-k)! (k-1)!} = n k \dbinom{n-1}{k-1}\\ & = n (k-1) \dbinom{n-1}{k-1} + n \dbinom{n-1}{k-1}\\ & = n(n-1) \dbinom{n-2}{k-2} + n \dbinom{n-1}{k-1} \end{align} Hence, \begin{align} \sum_{k = 0}^n k^2 \mathbb{P}(X=k) & = \sum_{k = 2}^n n(n-1) \dbinom{n-2}{k-2} p^k (1-p)^{n-k} + \sum_{k = 1}^n n \dbinom{n-1}{k-1} p^k (1-p)^{n-k}\\ & = n(n-1)p^2 + n p \end{align} Hence, \begin{align} \text{Var}(X) & = \sum_{k = 0}^n k^2 \mathbb{P}(X=k) - \left(\sum_{k = 0}^n k \mathbb{P}(X=k) \right)^2\\ & = n(n-1)p^2 + n p - (np)^2 = n^2p^2 - np^2 + np - n^2 p^2\\ & = np(1-p) \end{align}