Answer
There are two ways to solve this question. The first method is using prepositional algebra and another is using logic. In method 1, we use laws of prepositional logic to show that conjunction of all premises ($p_1$, $p_2$, $p_3$ ...... $p_n$) leads to the conclusion $p \rightarrow r$. Method 2 we use logic ( basically using English sentences to prove the same).
Work Step by Step
Method 1:
Given that : $p_1$, $p_2$, $p_3$ ...... $p_n$, $q$ premises leads to conclusion $r$, hence,
$(p_1 \land p_2 \land p_3 \land ......\land p_n \land q) \rightarrow r $
$\lnot (p_1 \land p_2 \land p_3 \land ......\land p_n \land q) \lor r$ {By equivalence of biconditionals}
$\lnot p_1 \lor \lnot p_2 \lor \lnot p_3 \lor ...... \lor \lnot p_n \lor \lnot q \lor r$ {De Morgan's Law}
$ (\lnot p_1 \lor \lnot p_2 \lor \lnot p_3 \lor ...... \lor \lnot p_n) \lor (\lnot q \lor r)$ {Associative Law for disjunctions}
$\lnot (p_1 \land p_2 \land p_3 \land ......\land p_n ) \lor (\lnot q \lor r)$ {De Morgan's Law}
$\lnot (p_1 \land p_2 \land p_3 \land ......\land p_n ) \lor (q \rightarrow r)$ {Equivalence of biconditionals)
$(p_1 \land p_2 \land p_3 \land ......\land p_n ) \rightarrow (q \rightarrow r)$ {Equivalence of biconditionals}
Hence, by this we can conclude that premises ($p_1$, $p_2$, $p_3$ ...... $p_n$) leads to the conclusion $p \rightarrow r$
Method 2:
Suppose $p_1$, $p_2$, $p_3$ ...... $p_n$ premises are true, then $r$ will be true when $q$ is true, because premises $p_1$, $p_2$, $p_3$ ...... $p_n$, $q$ leads to conclusion $r$ (which is already given). Now we can conclude that premises $p_1$, $p_2$, $p_3$ ...... $p_n$ leads to $p \rightarrow r$
