We consider 3 cases.

1) Suppose that $\deg Q = 1$. By looking at the leading coefficient, we have $Q(x) = \pm x + a$. If $Q(x) = x + a$, we have
$$P(x) = (x - (a+1))(x - (a+2)) \ldots (x - (a+9)).$$
And if $Q(x) = -x + a$ we have
$$P(x) = (a-1-x)(a-2-x) \ldots (a-9-x).$$

2) Suppose that $\deg P = 1$. By looking at the leading coefficient, we have $P(x) = \pm x + a$. In this case
$$Q(x) = \pm(x-1)(x-2) \ldots (x-9) - a.$$

3) Suppose $\deg P, \deg Q > 1$. Then, the only option is $\deg P = \deg Q = 3$.
Since $1,2, \ldots, 9$ are 9 roots of $P(Q(x))$, we have $P(x)$ have 3 roots $a, b, c$. For each root, for example $a$, the equation $Q(x) = a$ has 3 roots $a_1, a_2, a_3$ then $a_1, a_2, a_3, b_1, b_2, b_3, c_1, c_2, c_3$ form a permutation of $1,2, \ldots, 9$.
By Vieta's formula, we have
$$a_1 + a_2 + a_3 = b_1 + b_2 + b_3 \text{ and } a_1 a_2 + a_2 a_3 + a_3 a_1 = b_1 b_2 + b_2 b_3 + b_3 b_1$$
so $a_1^2 + a_2^2 + a_3^2 = b_1^2 + b_2^2 + b_3^2$. Since $x^2 \equiv 0$ or $1 \pmod{4}$ so the number of odd numbers in $a_1, a_2, a_3$ is the same as the number of odd numbers in $b_1, b_2, b_3$.
This implies that the number of odd numbers in $1,2, \ldots, 9$ is divisible by 3, which is a contradiction.