8. For a four-digit number $\overline{a b c d}(1 \leqslant a \leqslant 9,0 \leqslant b, c, d \leqslant 9)$, if $a>b, bd$, then $\overline{a b c d}$ is called a $P$ class number; if $ac, c<d$, then $\overline{a b c d}$ is called a $Q$ class number. Let $N(P)$ and $N(Q)$ represent the number of $P$ class numbers and $Q$ class numbers, respectively, then the value of $N(P)-N(Q)$ is $\qquad$.
When $d \neq 0$, let $\overline{a b c d}$ be a $P$ class number, then $\overline{d c b a}$ is a $Q$ class number, and vice versa. Thus, $N(P)-N(Q)$ is the number of $P$ class numbers that satisfy $d=0$, denoted by $A$ as the set of $P$ class numbers that satisfy $d=0$. Let $\overline{a b c d} \in A$, when $a=i, i \in\{1,2, \cdots, 9\}$, if $b=k, k \in\{0,1, \cdots, i-1\}$, then $c \in\{k+1, k+2, \cdots, 9\}$ has $9-k$ possibilities.
Therefore, $|A|=\sum_{i=1}^{9} \sum_{k=0}^{i-1}(9-k)=\sum_{i=1}^{9} \frac{i(19-i)}{2}=\frac{1}{2} \sum_{i=1}^{9}\left(19 i-i^{2}\right)=\frac{19}{2} \sum_{i=1}^{9} i-\frac{1}{2} \sum_{i=1}^{9} i^{2}$ $=\frac{19 \times 45}{2}-\frac{9 \times 10 \times 19}{12}=19 \times 15=285$. That is, $N(P)-N(Q)=285$.