Example of tree with > 6 vertices, tree would have depth = n after splay() deepest vertex
This is a good question whose answer can help us understand how splay tree tends to keep away from being imbalanced.

No, it is impossible for any rooted tree with more than 5 vertices to become a linear tree after operation splay() one of the deepest vertices, i.e. its depth is one less than the number of vertices.

Why?

For the sake of contradiction, let there be a rooted tree with more than 5 vertices which becomes a linear tree after splaying $x$, one of its deepest vertices. Consider the point of time just before the last splay step that moves $x$ to the root. There are three cases.


That last step is a zig-step (the following graph) or its mirroring zag-step.  Since the resulted tree is linear, part $A$ and $B$ are empty. That means, the left subtree of the root before that step contains node $x$ only. Since $x$ is the deepest node, the whole tree has at most 3 nodes, which is not true.





That last step is zig-zig step (the following graph) or its mirroring zag-zag.  Since the resulted tree is linear, part $A$, $B$ and $C$ are empty. That means, the left subtree of the root before that step contains node $P$ and $x$ only. Part $D$ can have at most two nodes; otherwise, either $x$ was not the deepest node or the resulted tree is not linear. So the whole tree has at most 5 nodes, which is not true.





That last step is a zig-zag step (the following graph) or its mirroring zag-zig. However, after that step, the root will alway has two children, which is not ture.






Exercise. Draw a tree with 5 vertices that becomes a linear tree after splaying one of its deepest vertices.