How to know the ionization power of an unknown metal? - difficult
I am guessing you know when a metal bar is becoming thinner, the metal is being oxidised, i.e $$ \ce{M_{(s)}} \rightarrow \ce{M^{n+}_{(aq)}}$$

$P^{p+}, Q^{q+}, s^{s+}, R^{r+}$
Let's start with beaker 3 and beaker 4

In beaker 3, you have $\ce{S_{(s)}}$ in $\ce{R^{r+}_{(aq)}}$, and you observer the metal bar becomes thinner. Which means, $\ce{S_{(s)}}$ is being oxidised by $\ce{R^{r+}_{(aq)}}$, or in other words, the reduction potential of $\ce{R^{r+}_{(aq)}}$ is greater than the reduction potential (oxidising power) of $\ce{S^{s+}_{(aq)}}$ 

$$\ce{R^{r+}_{(aq)}} &gt; \ce{S^{s+}_{(aq)}}$$ 

In beaker 4, you have $\ce{Q_{(s)}}$ in $\ce{S^{s+}_{(aq)}}$, and again the same reasoning applies. 

$\ce{Q_{(s)}}$ is being oxidised by $\ce{S^{s+}_{(aq)}}$, or in other words, the reduction potential of $\ce{S^{s+}_{(aq)}}$ is greater than the reduction potential (oxidising power) of $\ce{Q^{q+}_{(aq)}}$ 

$$\ce{S^{s+}_{(aq)}} &gt; \ce{Q^{q+}_{(aq)}}$$ 

Thus, so far we have: 

$$\ce{R^{r+}_{(aq)}} &gt; \ce{S^{s+}_{(aq)}} &gt; \ce{Q^{q+}_{(aq)}}$$ 

Let's look at beaker 1 and 2

In beaker 1 you have $\ce{P_{(s)}}$ in $\ce{Q^{q+}_{(aq)}}$, and oxidation doesn't take place. Which means the reduction potential of of $\ce{Q^{q+}_{(aq)}}$ is less than that of $\ce{P^{p+}_{(aq)}}$

$$\ce{P^{p+}_{(aq)}} &gt; \ce{Q^{q+}_{(aq)}}$$ 

And similarly, in beaker 2 you have $\ce{P_{(s)}}$ in $\ce{R^{r+}_{(aq)}}$, and nothing happens. Which means that the Which means the reduction potential of of $\ce{R^{r+}_{(aq)}}$ is less than that of $\ce{P^{p+}_{(aq)}}$

$$\ce{P^{p+}_{(aq)}} &gt; \ce{R^{r+}_{(aq)}}$$ 

So the final order becomes (in terms of reduction potential, or oxidising ability):

$$ \ce{P^{p+}_{(aq)}} &gt;\ce{R^{r+}_{(aq)}} &gt; \ce{S^{s+}_{(aq)}} &gt; \ce{Q^{q+}_{(aq)}}$$