= Solution
Take the five-element <partially ordered set> $P=\{\bot,a,b,u,v\}$. Its strict comparisons are $\bot<a$, $\bot<b$, and $a<u$, $a<v$, $b<u$, $b<v$, together with their transitive consequences. The elements $a,b$ are incomparable, as are $u,v$.
\Image[/past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2003/math-prelims-2003-completeness-counterexamples.png]
{title=Two different failures of order completeness}
A <finite directed set has a greatest element> whenever it is nonempty: list its elements and repeatedly use directedness to find an internal <upper bound> of the previously chosen element and the next one. The final element is above every member of the <directed set>, and is its <least upper bound>. Thus every nonempty <directed set> in $P$ has a <least upper bound>. The empty subset also has a <least upper bound>, namely $\bot$. Therefore $P$ satisfies the paper's completeness condition under either convention.
However, $\{a,b\}$ is bounded above by both $u$ and $v$. A <least upper bound> would have to be an <upper bound> lying below both $u$ and $v$. Neither $u$ nor $v$ has that property, and none of $\bot,a,b$ is an <upper bound> of $\{a,b\}$. Hence
$$
\boxed{P\text{ is directed-complete but not bounded-complete}.}
$$
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