, 2022

If $P$ is a constant, then $P \equiv 0$. If there exist $a, b$, such that $P(a) < 0$ and $P(b) > 0$, due to continuity $P$ will have a real root between $a$ and $b$. If no such $a, b$ exist, then $P$ attains either only positive or only negative values. WLOG $P(x) > 0$ for all $x$. If $a_i \neq a_{2022}$ for some $i$, due to the odd degree of $Q$, there exists $x_0$, satisfying $Q(x_0) = \frac{b_{2022}-b_i}{a_i-a_{2022}}$, which gives rise to
$$P(a_1Q(x_0) + b_1) + \dots + P(a_{i-1}Q(x_0) + b_{i-1}) + P(a_{i+1}Q(x_0) + b_{i+1}) + \dots + P(a_{2021}Q(x_0) + b_{2021}) = 0.$$

A contradiction with $P(x) > 0$. Hence $a_1 = a_2 = \dots = a_{2022} = a$, meaning that
$$P(aQ(x) + b_1) + \dots + P(aQ(x) + b_{2021}) = P(aQ(x) + b_{2022}).$$
Denote by $b$ and $c$ the leading coefficients of $P$ and $Q$, respectively, and let $d = \deg P$. The LHS leading coefficient equals $2021ba^d c^d$, while the RHS leading coefficient is $ba^d c^d$. Since $a, c \neq 0$, we deduce that $b = 0$, e.g., $P(x) \equiv 0$.