The given board, call it $D$, can be viewed as the difference of two (unsorted) Ferrers boards with $m=9$ columns each: $A=(4,5,6,7,8,9,8,7,6)$ and $B=(3,2,1,0,1,2,3,4,5)$. We view $B$ as a sub-board of $A$ and are interested in enumerating placements of $k$ non-attacking rooks in $A$ such that no rook appears in $B$. This can be computed via inclusion-exclusion as follows:
$$r_k(D) = \sum_{S\subseteq[m]\atop |S|=k} \sum_{T\subseteq S} (-1)^{|T|} r_k(A_{S\setminus T}\| B_T),$$
where $B_T$ is the sub-board of $B$ formed by the columns indexed by $S$, similarly $A_{S\setminus T}$ is the sub-board of $A$ formed by the columns indexed by the complement of $S\setminus T$, and $\|$ denotes concatenation of the two boards.
Here, each summand can be easily computed -- if $A_{S\setminus T}\| B_T$ is formed by elements $c_1\leq c_2\dots \leq c_k$, then
$$r_k(A_{S\setminus T}\| B_T) = \prod_{i=1}^k (c_i - i + 1).$$
So, the formula for $r_k(D)$ involves up to $3^m$ summands but depends on sorting.
Below we construct more explicit formula for the whole rook polynomial $R_D(x):=\sum_k r_k(D)x^k$, with the same number of summands.
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Let $A=(a_1,\dots,a_m)$ and $B=(b_1,\dots,b_m)$. Let $p_i$ be the indicator for $a_i\in S\setminus T$ and $q_i$ be the indicator for $b_i\in T$. It follows that $(p_i,q_i)\in\{(0,0),(1,0),(0,1)\}$. Let $\sigma$ be a sorted permutation of $A\|B$, and $\delta$ be obtained from $\sigma$ by replacing each $a_i$ and $b_i$ with $p_i$ and $q_i$, respectively. Finally, let $\tau_A(i)$ be the order number of $a_i$ in $\sigma$ (i.e., $\delta_{\tau_A(i)}=p_i$), and $\tau_B(i)$ be the order number of $b_i$ in $\sigma$ (i.e., $\delta_{\tau_B(i)}=q_i$). Then
$$(\star)\qquad R_D(x) = \sum_{(p,q)\in\{(0,0),(1,0),(0,1)\}^m} (-1)^{q_1+\cdots+q_m} \prod_{i=1}^m \big(1+p_i(A_i-\sum_{j=1}^{\tau_A(i)-1} \delta_j)x + q_i(B_i-\sum_{j=1}^{\tau_B(i)-1} \delta_j)x\big).$$
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**Example.** In the given example, we have $\sigma = (0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9)$ and $\delta = (q_4,q_3,q_5,q_2,q_6,q_1,q_7,p_1,q_8,p_2,q_9,p_3,p_9,p_4,p_8,p_5,p_7,p_6)$. Correspondingly, $\tau_A=(8,10,12,14,16,18,17,15,13)$ and $\tau_B=(6,4,2,1,3,5,7,9,11)$. This leads to the formula for the rook polynomial for the board $D$:
$$R_D(x) =
1 + 39 x + 563 x^2 + 3833 x^3 + 13039 x^4 + 21773 x^5 + 16516 x^6 + 4884 x^7 + 425 x^8 + 7 x^9.$$
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**ADDED.** We can eliminate dependency between $p_i$ and $q_i$ by slightly modifying the formula $(\star)$:
$$R_D(x) = \sum_{p,q\in\{0,1\}^m} (-1)^{q_1+\cdots+q_m} \prod_{i=1}^m \big(1-p_iq_i+p_i(1-q_i)(A_i-\sum_{j=1}^{\tau_A(i)-1} \delta_j)x + q_i(1-p_i)(B_i-\sum_{j=1}^{\tau_B(i)-1} \delta_j)x\big).$$
It can be seem that when $p_i=q_i=1$, the corresponding term becomes zero, while in all other cases, it is same as in $(\star)$.
Next, by substituting $p_i := \frac{1-y_i}2$ and $q_i := \frac{1-z_i}2$, we switch to summation over $y,z\in\{-1,1\}^m$. Let $\delta'$ be obtained from $\delta$ by such substitution. Then it can be seen that $R_D(x)$ equals the sum of the (polynomial in $x$) coefficients of terms $y^{\alpha}z^{\beta}$ in
$$F_D(x;y_1,\dots,y_m,z_1,\dots,z_m):=\prod_{i=1}^m \big(4-(1-y_i)(1-z_i)+(1-y_i)(1+z_i)(A_i-\sum_{j=1}^{\tau_A(i)-1} \delta'_j)x + (1+y_i)(1-z_i)(B_i-\sum_{j=1}^{\tau_B(i)-1} \delta'_j)x\big),$$
where all components of $\alpha$ are even, and all components of $\beta$ are odd. From computational perspective, it is worth to consider $F_D$ modulo $y_i^2-1$ and $z_i^2-1$, which has at most $2^{2m}$ terms. That is,
$$R_D(x) = [y_1^0\cdots y_m^0 z_1\cdots z_m]\ \big( F_D(x;y_1,\dots,y_m,z_1,\dots,z_m) \bmod (y_1^2-1,\dots,y_m^2-1,z_1^2-1,\dots,z_m^2-1)\big).$$
This formula may be not better than $(\star)$ from computational perspective, but it looks nice anyway.