Solution:

Fill the left block arbitrarily: $9!$ ways. Let $R_1, R_2, R_3$ be the sets of three digits in its rows. In the middle block, row $i$ must avoid $R_i$ (those digits already appear in row $i$), and the block's three rows must partition $\{1, \ldots, 9\}.$ Say its top row takes $j$ digits from $R_2$ and $3 - j$ from $R_3.$ Balancing the three rows then forces the middle row to take $3 - j$ digits from $R_1$ together with all $j$ remaining digits of $R_3,$ and the bottom row is determined. The number of content choices is $$\sum_{j=0}^{3} \binom{3}{j}\binom{3}{3-j}^2 = 1 + 27 + 27 + 1 = 56.$$

The right block's row contents are then forced (row $i$ takes whatever is missing from row $i$), and each of the six rows of the middle and right blocks can be ordered internally in $3!$ ways. The total is $$9! \cdot 56 \cdot 6^6 = (2^7 \cdot 3^4 \cdot 5 \cdot 7)(2^3 \cdot 7)(2^6 \cdot 3^6) = 2^{16} \cdot 3^{10} \cdot 5^1 \cdot 7^2.$$

Therefore $p \cdot a + q \cdot b + r \cdot c + s \cdot d = 2 \cdot 16 + 3 \cdot 10 + 5 \cdot 1 + 7 \cdot 2 = 81.$