The pre-order traversal of a binary search tree is given by 12, 8, 6, 2, 7, 9, 10, 16, 15, 19, 17, 20. Then the post-order traversal of this tree is:

For a Binary Search Tree (BST), the in-order traversal is always in ascending sorted order.

Given the pre-order traversal: Pre-order: $12, 8, 6, 2, 7, 9, 10, 16, 15, 19, 17, 20$

Sorting the elements gives the in-order traversal: In-order: $2, 6, 7, 8, 9, 10, 12, 15, 16, 17, 19, 20$

1. The root of the tree is the first element of the pre-order traversal: 12.

2. In the in-order traversal, elements to the left of $12$ are $\{2, 6, 7, 8, 9, 10\}$ (Left Subtree) and to the right are $\{15, 16, 17, 19, 20\}$ (Right Subtree).

3. Reconstructing the Left Subtree (Root = $8$):

   • Left of $8$: $\{2, 6, 7\}$ (Pre-order: $6, 2, 7$; Root = $6$, Left child = $2$, Right child = $7$). Post-order: $2, 7, 6$.

   • Right of $8$: $\{9, 10\}$ (Pre-order: $9, 10$; Root = $9$, Right child = $10$). Post-order: $10, 9$.

   • Post-order of left subtree: $2, 7, 6, 10, 9, 8$.

4. Reconstructing the Right Subtree (Root = $16$):

   • Left of $16$: $\{15\}$. Post-order: $15$.

   • Right of $16$: $\{17, 19, 20\}$ (Pre-order: $19, 17, 20$; Root = $19$, Left child = $17$, Right child = $20$). Post-order: $17, 20, 19$.

   • Post-order of right subtree: $15, 17, 20, 19, 16$.

Combining Left Subtree, Right Subtree, and Root gives the overall post-order traversal: $2, 7, 6, 10, 9, 8, 15, 17, 20, 19, 16, 12$.

Therefore, option B is correct.