How many arrangements? $(a, b, c, d, x, y, z, w)$

We are counting strictly order-preserving maps $P\to[8]$, where $[8]$ denotes the poset $\{1<2<\dots<8\}$ and $P=[2]\times[4]$ with $(a,b)\le(c,d)$ if and only if $a\le c$ and $b\le d$: $$ P : \quad \begin{array}{ccccccc} (1,1) & < & (1,2) & < & (1,3) & < & (1,4) \\ \style{display: inline-block; transform: rotate(90deg)}{<} && \style{display: inline-block; transform: rotate(90deg)}{<} && \style{display: inline-block; transform: rotate(90deg)}{<} && \style{display: inline-block; transform: rotate(90deg)}{<} \\ (2,1) & < & (2,2) & < & (2,3) & < & (2,4). \end{array} $$ Hence, the number we are looking for is $\Omega^\circ_P(4)$, where $\Omega^\circ_P$ is the strict order polynomial of $P$.

For the strict order polynomial, there is the formula $$ \Omega_P(n) = \sum_{w\in\mathcal L(P)} \binom{n+\operatorname{des}(w)}{|P|}, $$ where $\mathcal L(P)$ is the set of linear extensions of $P$ and $\operatorname{des}(w)$ is the number of descents of the linear extension $w$ with respect to some fixed natural labeling of $P$.

In our case, we can pick the natural labeling $\begin{matrix}1234\\5678\end{matrix}$. As mentioned by @saulspatz, there are $14$ linear extensions of $P$ by the hook length formula for counting standard Young tableaux. A list of these with their descent numbers is: $$ 12345678, d=0,\\ 12354678, d=1,\\ 12534678, d=1,\\ 15234678, d=1,\\ 12356478, d=1,\\ 12536478, d=2,\\ 15236478, d=2,\\ 12563478, d=1,\\ 15263478, d=2,\\ 12356748, d=1,\\ 12536748, d=2,\\ 15236748, d=2,\\ 12563748, d=2,\\ 15263748, d=3. $$ Here a descent is any position where a number is followed by a smaller number, for example in $15263478$ the descents are $52$ and $63$.

Hence, we have $$\Omega^\circ_P(n) = \binom{n}{8} + 6 \binom{n+1}{8} + 6 \binom{n+2}{8} + \binom{n+3}{8}$$

This finally yields $$\Omega^\circ_P(8) = \binom 8 8 + 6 \binom 9 8 + 6 \binom {10} 8 + \binom {11} 8 = 490.$$