The answer is
7 ______ 934 ) 7210 6538 ---- 672 (H, I, J, K, L, M, N, O, P, Q) = (9, 3, 4, 7, 2, 1, 0, 6, 5, 8)
Some reasoning (for the equations, $KL$ denotes digit concatenation; all multiplications use explicit $\times$):
$L \ne 0$, since otherwise $N=Q$ from the lowest digit. Let's say $KL + IQ = bMN$ where $b\in\{0,1\}$. Then $OP + O + b = KL$. We know that $K = O + 1$ and $L \ge 1$, so $$OP + O + b = 10\times(K-1) + P + O + b = KL = 10\times K + L \\ P + O + b = L + 10 \\ P + O = L + 10 - b \ge 10$$ If we plug in various values of $K$ into $HIJ \times K > OP00$, we get $$HIJ \times 2 > 1900, HIJ \times 3 > 2800, \dots, HIJ \times 8 > 7300$$ and in all cases, $H=9$ and $I\ge 1$.
Now let's say $IJ \times K = xIQ$ where $1 \le x \le I$. ($x$ cannot be 0 because $I$ is nonzero and $K > 1$.) Then we can say $OP = H \times K + x = 9 \times K + x$. If we plug it back into $OP + O + b = KL$, we get $$9\times K + x + (K - 1) + b = KL \\ L = x + b - 1$$
At this point, let's assume $b=0$. As Ross Millikan pointed out (and we found $I\ne 0$), the zero must be either $M$ or $N$. But if $b=0$, $KL + IQ = MN$, so $M\ne 0$ and subsequently $N=0$. We can find out the following: $$L = x-1 \\ KL + IQ = M0 \Rightarrow L+Q=10, K+I+1=M \\ L = x-1 \Rightarrow x+Q=11 \\ Q \le 8 \Rightarrow x \ge 3 \\ M \le 8 \Rightarrow K+I \le 7 $$ Now, $IJ \times K = xIQ > 300$, but any choice of $K$ and $I$ subject to $K+I \le 7$ cannot satisfy $IJ \times K > 300$. Therefore $b=1$.
Plugging in $b=1$ to $L=x+b-1$, we simply get $L=x$, and we can write $IJ \times K = LIQ$. Under the constraint that no digits in the equation can be 0 or 9 and the five digits $JKLIQ$ are unique, I decided to brute-force the equation and found four candidates: $$34 \times 7 = 238, 67 \times 4 = 268, 72 \times 8 = 576, 83 \times 7 = 581$$ With two additional conditions that $O=K-1$ and $P=10-K+L$ are also unique, only the first one survives, and gives the final answer.