Step 1: Write down exactly which symbols appear in the rule.
The rule for specialising on P1 is $\dfrac{E_1}{S_1+h_1} > \dfrac{E_2}{h_2}$. The only symbols in this expression are $E_1$, $S_1$, $h_1$, $E_2$, and $h_2$. Notice that $S_2$, the search time for P2, is simply not part of the formula at all.
Step 2: Translate abundance into the symbols that matter.
We are told search time and abundance move in opposite directions, more abundant prey means shorter search time. So the abundance of P1 controls $S_1$, and the abundance of P2 would control $S_2$, if $S_2$ appeared anywhere. Since $S_2$ is absent from the rule, the abundance of P2 cannot influence this decision through the rule at all.
Step 3: Check each option against the symbol list.
Handling time of P1 is $h_1$, which is present, sitting with $S_1$ in the left-hand denominator, so option (A) matters. Handling time of P2 is $h_2$, present on the right-hand side, so option (B) matters. Abundance of P1 sets $S_1$, present on the left, so option (C) matters. Abundance of P2 would set $S_2$, which never appears, so option (D) does not matter.
Step 4: Interpret why P2's abundance drops out biologically.
A forager specialising on P1 still runs into P2 items while searching for P1, it just chooses to ignore them. So the rate at which it finds P2, which depends on P2's abundance, never enters the calculation, only how long it takes to process a P2 item once accepted, which is $h_2$.
Step 5: Conclude.
Handling time of P1, handling time of P2, and abundance of P1 all feed into the decision rule; abundance of P2 does not.
\[ \boxed{\text{(A), (B), (C)}} \]