Comparing coefficients of boolean functions
Consider the function $P(x_1) = x_1$. Then $P(0) = 0$, $P(1) = 1$, and the polynomial representation is $x_1$, but your formula gives $c_0 = 1$ and $c_1 = -1$. However, if we define $\hat{P}(b) = 2^{-n} \sum_{a \in \{0,1\}^n} (-1)^{\langle a,b \rangle} P(a)$ then
$$ P(x) = \sum_{b \in \{0,1\}^n} \hat{P}(b) (-1)^{\langle x,b \rangle}. $$
This is known as the Fourier expansion of $P$. To relate this to your expansion, notice that
$$ (-1)^{\langle x,b \rangle} = \prod_{i\colon b_i=1} (-1)^{x_i} = \prod_{i\colon b_i=1} (1-2x_i). $$
It is now an exercise to relate the coefficients $\hat{P}(b)$ to the coefficients $c_b$ you are interested in.