8. For a four-digit number $\overline{a b c d}(1 \leqslant a \leqslant 9,0 \leqslant b 、 c 、 d \leqslant$ $9)$, if $a>b, b<c, c>d$, then $\overline{a b c d}$ is called a $P$ class number; if $a<c, c<d$, 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 $\qquad$
8. 285.

Let the sets of all P-type numbers and Q-type numbers be denoted as $A$ and $B$, respectively. Let the set of all P-type numbers whose unit digit is zero be denoted as $A_{0}$, and the set of all P-type numbers whose unit digit is not zero be denoted as $A_{1}$.

For any four-digit number $\overline{a b c d} \in A_{1}$, map it to the four-digit number $\overline{d c b a}$.
Note that, $a > b$, and $bd \geqslant 1$.
Then, $\overline{d c b a} \in B$.
Conversely, each $\overline{d c b a} \in B$ uniquely corresponds to an element $\overline{a b c d} \in A_{1}$.
Therefore, a one-to-one correspondence is established between $A_{1}$ and $B$.
Hence, $N(P) - N(Q) = |A| - |B|$
$= |A_{0}| + |A_{1}| - |B| = |A_{0}|$.
Next, we calculate $|A_{0}|$.
For any four-digit number $\overline{a b c 0} \in A_{0}$, $b$ can take values $0, 1, \cdots, 9$. For each $b$, given $b < a \leqslant 9$ and $b < c \leqslant 9$, $a$ and $c$ each have $9 - b$ possible values.
Thus, $|A_{0}| = \sum_{b=0}^{9} (9 - b)^{2} = \sum_{k=1}^{9} k^{2} = 285$.
Therefore, $N(P) - N(Q) = 285$.