Observe that $U(f,\mathcal P)-L(f,\mathcal P)=3$. So, the partition you are looking for is preferably finer than the given partition $\cal P$. So, adding more points to $\cal P$ to form the partition \cal P' is a way to go!
Why should this work? Let $P$ and $Q$ be two partitions of $[a,b]$. Suppose $P$ is finer than $Q$. (By slight abuse of notation, most authors also write $P \supseteq Q$.) Then, $L(f,Q) \le L(f,P) \\U(f,Q) \ge U(f,P)$
So, the inequalities would imply, U(f,\mathcal P')-L(f,\mathcal P') \le U(f,\mathcal P)-L(f, \mathcal P)
The slack inequality we show above does not prove that there actually exists a partition which achieves the upper bound that the question claims.
So, note that $f$ is monotonic and hence $f \in \scr R [a,b]$. So, $\lim_{\|P\| \to 0}U(f,P)-L(f,P)=0$
Given an $\varepsilon \gt 0,$ there exists at least one partition $P_ \varepsilon$ such that $U(P_\varepsilon,f)-L(P_\varepsilon, f) \lt \varepsilon$ Now, we are done showing the existence of a partition. Re-interpreting that limit, we must have that refining the partition, must eventually lead to a decrease of $U(f,P)-L(f,P)$ by any given positive quantity.
So, we must have a partition as sought by the question.
I'll only remark that $P_2=\left\{1,\dfrac 3 2, \dfrac 8 5, 2, \dfrac 5 2, 3\right\}$ works good enough, while I'll leave the details for you.
The general philosophy is, refining partitions increases the lower sum while it decreases the upper sum. It is what one would expect, if the function is integrable. From high school, we know that making finer "strips" will eventually give you the area under the curve.
Note: I am not claiming that it is the way to go, but one of the ways clearly seen given $\cal P$.