Some authors will say that the (p∧¬p) consists of a contradiction, while having p as well as ¬p is not a contradiction. One definition of a contradiction comes as a formula F such that for all possible truth values of F's atomic propositions, F has truth value of false. Under this definition, having p as well as ¬p is not a contradiction, and you can only infer a contradiction from having p as well as ¬p if you have a conjunction introduction rule, or something equivalent to such a rule.
Not all mathematicians use a framework where ((p∧¬p)$\implies$q) will hold. You might want to see the applications of relevance logic here. Mitch points out here that ((p∧¬p)$\implies$q) fails for relevance logic.
Though we can fairly easily guess something about the formation rules of your logical system, we know nothing about the semantics of your logical system, nor anything about the rules of inference of your system, nor anything about the axioms of such a system. Since we don't know the rules of inference and don't know the axioms of your system, we don't know what qualifies as a formal proof and what doesn't qualify as a formal proof for your logical system.
If we had a completeness meta-theorem (if a formula qualifies as a tautology, then it qualifies as a theorem), and knew the semantics of your logical system (the truth values that can get meaningfully assigned to propositions), then we could know what we could prove for your logical system, just from calculations with truth values. We just verify that some formula qualifies as a tautology via truth-value calculations, and then we can invoke the respective completeness meta-theorem which implies all tautologies as theorems. This all presupposes the rules of inference and axioms your logical system adequate with respect to the semantics of your system, whatever they might consist of. But, even given just a completeness meta-theorem, since we know nothing about the semantics of your system, and we have nothing which suggests anything about the semantics of your logical system, we don't even know if a given formula comes as provable or not. Not all logical systems have the same tautologies even those that do have a completeness meta-theorem, let alone the same theorems, since the semantics of logical systems does differ significantly.
Your proof comes as not even wrong, since we have no idea of what consists of a proof for your system.
The statement "((p$\land$$\lnot$p)$\implies$ q)" also comes as not even wrong, since we can't even tell if it consists of a tautology, nor can we tell if it qualifies as a theorem for what you want to talk about.
If you presuppose classical logic with its two truth values, and presuppose that your system has a completeness meta-theorem, in an informal context, you really only need to proceed as follows:
- So, ((P$\land$$\lnot$P)$\implies$Q)=1.
- Completeness Meta-Theorem
- Therefore, ((P$\land$$\lnot$P)$\implies$Q) is a theorem.
Really, this proof should work for any adequate system of logic where (0$\implies$P)=1, (P$\land$$\lnot$P)=0, which has a completeness meta-theorem also. But, not all logical systems have this going on, and since we have no idea about your rules of inference and your axioms, nor any idea about your semantics, your proof is not correct in a certain sense where "not correct" just means "other than correct" rather than "wrong" necessarily. In truth, your proof is not wrong either, since we have no idea what right and wrong mean in your context. To repeat, your proof is not even wrong.