= Solution
Replace the one-place marker $\mathrm{Empty}$ by three nullary <planning fluents> $q_0,q_1,q_2$, meaning that zero, one or two items are aboard. Initially only $q_0$ is true. Exactly one of these facts is maintained. No additional object constants are needed.
For compactness, define the common loading <preconditions>
$$
L(x,s)=\{\mathrm{isItem}(x),\mathrm{isBank}(s),
\mathrm{at}(x,s),\mathrm{at}(\mathrm{BOAT},s)\}
$$
and common unloading <preconditions>
$$
U(x,s)=\{\mathrm{isItem}(x),\mathrm{isBank}(s),
\mathrm{onBoat}(x),\mathrm{at}(\mathrm{BOAT},s)\}.
$$
The four <STRIPS> actions are
$$
\begin{aligned}
\mathrm{LoadFirst}(x,s):\quad
P&=L(x,s)\cup\{q_0\},\\
A&=\{\mathrm{onBoat}(x),q_1\},&
D&=\{\mathrm{at}(x,s),q_0\};\\[3pt]
\mathrm{LoadLast}(x,s):\quad
P&=L(x,s)\cup\{q_1\},\\
A&=\{\mathrm{onBoat}(x),q_2\},&
D&=\{\mathrm{at}(x,s),q_1\};\\[3pt]
\mathrm{UnloadFirst}(x,s):\quad
P&=U(x,s)\cup\{q_2\},\\
A&=\{\mathrm{at}(x,s),q_1\},&
D&=\{\mathrm{onBoat}(x),q_2\};\\[3pt]
\mathrm{UnloadLast}(x,s):\quad
P&=U(x,s)\cup\{q_1\},\\
A&=\{\mathrm{at}(x,s),q_0\},&
D&=\{\mathrm{onBoat}(x),q_1\}.
\end{aligned}
$$
Here the first unloading removes one of two items, and the last unloading removes the sole remaining item; items do not require fixed seat identities. The movement action remains as repaired in part (a) and changes none of the occupancy facts.
The <planning state invariant> is $q_j$ true exactly when $j$ items are aboard. Every action changes the marker by the same amount as the actual onboard count, so it follows by <mathematical induction> from the initial empty boat. In particular, a third load is impossible.
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