= Solution
The <splitting field> of $f\in K[t]$ is the smallest extension $L/K$ in which $f$ is a product of linear factors; equivalently, $L$ is generated over $K$ by all its <roots of a polynomial>. To prove <existence and uniqueness of splitting fields>, adjoin a root of an <irreducible polynomial> dividing $f$ and repeat until $f$ splits. If $L$ and $L'$ are two resulting fields, an embedding between the fields generated so far extends by sending each newly adjoined root to a root of its transformed <minimal polynomial>. Repetition gives a $K$-embedding $L\to L'$, whose image contains every root of $f$ and is therefore all of $L'$. Thus the splitting field exists and is unique up to $K$-isomorphism.
Now write $K=\mathbb F_q$, and let the distinct irreducible factors of $f$ have degrees $d_1,\ldots,d_r$. The <irreducible factors of a finite-field Frobenius polynomial> show that a root of the degree-$d_i$ factor lies in $\mathbb F_{q^n}$ exactly when $d_i\mid n$. Hence the <splitting field over a finite field> is
$$
L=\mathbb F_{q^m},
\qquad
\boxed{[L:K]=m=\operatorname{lcm}(d_1,\ldots,d_r)}.
$$
Every <finite field> is a <perfect field>, so $L/K$ is separable; it is normal because it is a splitting field. It is consequently a <Galois extension>. More explicitly, the <Galois group of a finite field extension> is the <cyclic group> of order $m$ generated by the <finite-field Frobenius automorphism> $x\mapsto x^q$.
Finally, let $N_2(n)$ denote the <number of monic irreducible polynomials over a finite field> of degree $n$ over $\mathbb F_2$. Since
$$
N_q(n)=\frac1n\sum_{d\mid n}\mu(d)q^{n/d},
$$
and $\mu(1)=1$, $\mu(3)=-1$, $\mu(9)=0$, we obtain
$$
\boxed{|\mathcal P_9|=N_2(9)=\frac{2^9-2^3}{9}=56.}
$$
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