For $0 \lt x \lt 1,$ $$\begin{aligned} P(x) &= (1-x^{2310})^6 \\ &\quad {}\cdot \frac{1}{(1-x^{105})(1-x^{70})} \\ &\quad {}\cdot \frac{1}{(1-x^{42})(1-x^{30})}, \end{aligned}$$ and each factor $\frac{1}{1 - x^k}$ expands as a geometric series. Since $2022 \lt 2310,$ the factor $(1 - x^{2310})^6$ contributes only its constant term $1,$ so the coefficient of $x^{2022}$ is the number of nonnegative integer solutions of $105a + 70b + 42c + 30d = 2022.$

Reducing modulo $2$ gives $105a \equiv 2022,$ so $a$ is even; modulo $3$ gives $70b \equiv 2022 \equiv 0,$ so $3 \mid b;$ modulo $5$ gives $2c \equiv 2022 \equiv 2,$ so $c \equiv 1 \pmod 5;$ modulo $7$ gives $2d \equiv 2022 \equiv 6,$ so $d \equiv 3 \pmod 7.$ Writing $a = 2a',$ $b = 3b',$ $c = 5c' + 1,$ $d = 7d' + 3$ turns the equation into $$\begin{aligned} &210(a' + b' + c' + d') + 42 \\ &\quad {}+ 90 = 2022, \end{aligned}$$ so $$a' + b' + c' + d' = 9.$$

By stars and bars there are $\binom{12}{3} = 220$ solutions, so the coefficient is $220.$