Solution:
Again, we have
$$p(x)=\min _{0 \leq k \leq n}\left\{a_{k}+k x\right\} .$$
So the graph of $y=p(x)$ can be drawn as follows: first, draw all the lines $y=a_{k}+k x$, $k=0,1, \ldots, n$, then trace out the lowest broken line, which then is the graph of $y=p(x)$.
So $p(x)$ is piecewise linear and continuous, and has slopes from the set $\{0,1,2, \ldots, n\}$. We know from the previous problem that $p(x)$ is concave, and so its slope must be decreasing (this can also be observed simply from the drawing of the graph of $y=p(x)$ ). Then, let $r_{k}$ denote the $x$-coordinate of the leftmost kink such that the slope of the graph is less than $k$ to the right of this kink. Then, $r_{n} \leq r_{n-1} \leq \cdots \leq r_{1}$, and for $r_{k-1} \leq x \leq r_{k}$, the graph of $p$ is linear with slope $k$. Note that it is possible that $r_{k-1}=r_{k}$, if no segment of $p$ has slope $k$. Also, since $a_{n} \neq \infty$, the leftmost piece of $p(x)$ must have slope $n$, and thus $r_{n}$ exists, and thus all $r_{i}$ exist.
Now, compare $p(x)$ with
$$\begin{aligned} q(x) & =a_{n} \odot\left(x \oplus r_{1}\right) \odot\left(x \oplus r_{2}\right) \odot \cdots \odot\left(x \oplus r_{n}\right) \\ & =a_{n}+\min \left(x, r_{1}\right)+\min \left(x, r_{2}\right)+\cdots+\min \left(x, r_{n}\right) . \end{aligned}$$
For $r_{k-1} \leq x \leq r_{k}$, the slope of $q(x)$ is $k$, and for $x \leq r_{n}$ the slope of $q$ is $n$ and for $x \geq r_{1}$ the slope of $q$ is 0. So $q$ is piecewise linear, and of course it is continuous. It follows that the graph of $q$ coincides with that of $p$ up to a translation. By taking any $x<r_{n}$, we see that $q(x)=a_{n}+n x=p(x)$, we see that the graphs of $p$ and $q$ coincide, and thus they must be the same function.