Step 1: Identify the root from preorder.
In preorder traversal, the first element is always the root.
Thus, root \( = P \).
Step 2: Split inorder traversal around root.
In inorder:
\[
S, Q, E, P, F, R, G
\]
Elements left of \( P \) form left subtree:
\[
S, Q, E
\]
Elements right of \( P \) form right subtree:
\[
F, R, G
\]
Step 3: Construct left subtree.
From preorder after removing \( P \):
\[
Q, S, E, R, F, G
\]
Left subtree preorder:
\[
Q, S, E
\]
Root of left subtree \( = Q \).
In inorder left part:
\[
S, Q, E
\]
Left of \( Q \): \( S \).
Right of \( Q \): \( E \).
Thus left subtree:
\[
\text{ Q}
\]
\[
\text{ / \textbackslash}
\]
\[
\text{ S E}
\]
Step 4: Construct right subtree.
Remaining preorder:
\[
R, F, G
\]
Root \( = R \).
In inorder right part:
\[
F, R, G
\]
Left of \( R \): \( F \).
Right of \( R \): \( G \).
Step 5: Write postorder traversal.
Postorder = Left subtree, Right subtree, Root.
Left subtree postorder:
\[
S, E, Q
\]
Right subtree postorder:
\[
F, G, R
\]
Whole tree postorder:
\[
S, E, Q, F, G, R, P
\]
Thus option (B) is correct.