Naive Bayes Classifier assumes that attributes have independent distribution.
So, if there are $K$ features in the dataset as $x_1,x_2,...,x_K$ and the class label is $y$.
Here, both $x$ and $y$ are boolean variables. It means $x_i=0$ or $x_i=1$ for $1\leq i \leq K$ and $y=0$ or $y=1$
Now, according to the "Independent" assumption of Naive Bayes Classifier,
$\mathbb{P}(x_1,x_2,x_3,...,x_K \ | y)=\mathbb{P}(x_1 \ | \ y)\mathbb{P}(x_2 \ | \ y)\mathbb{P}(x_3 \ | \ y)...\mathbb{P}(x_K \ | \ y)$ $\; \; \; (1)$
Now, since, Bayesian Classifiers use the Bayes theorem i.e. $P(y|x)=\frac{P(x|y)P(y)}{P(x)}$ but we ignore the $P(x)$ here because it is same for all the classes "$y$" and so it is irrelavant here.
So,
$\mathbb{P}(y|x_1,x_2,x_3,...,x_K)=\mathbb{P}(x_1,x_2,x_3,...,x_K \ | y)\mathbb{P}(y)$
Now, according to the equation $(1),$
$\mathbb{P}(y|x_1,x_2,x_3,...,x_K)=\mathbb{P}(x_1 \ | \ y)\mathbb{P}(x_2 \ | \ y)\mathbb{P}(x_3 \ | \ y)...\mathbb{P}(x_K \ | \ y) \mathbb{P}(y)$
Now, to estimate each $\mathbb{P}(x_i \ | \ y),$ we have to find $2$ values for each $y=0$ and $y=1$.
For example, if we have found $\mathbb{P}(x_i=1 \ | \ y=0)$ then we can easily find $\mathbb{P}(x_i=0 \ | \ y=0)= 1- \mathbb{P}(x_i=1 \ | \ y=0)$ and vice-versa.
The reason is being they are independent features.
Same goes for $y=1.$
Hence, we need $2$ values for estimating each $\mathbb{P}(x_i \ | \ y)$ and so we need total $2+2+...K \ times = 2K$ values and we also need one value to estimate $\mathbb{P}(y)$.
So, we need total $2K+1$ parameters.
$\textbf{Note:}$
If the features are not independent then we need total $2^{K+1}-2$ parameters to estimate $\mathbb{P}(x_1,x_2,x_3,...,x_K \ | y)$ because for $y=0$, you will need $2^K-1$ values because you can compute remaining one value by making probability sum as 1 and similarly for $y=1$, you will need $2^K-1$ values.
Hence total you will need $2(2^K-1)=2^{K+1}-2$ to estimate $\mathbb{P}(x_1,x_2,x_3,...,x_K \ | y).$
