Bram28's natural deduction proof shows that it is valid. However, suppose it were not valid. One way to check if it were invalid is to try to find a counter example that would make all of the premises true but the conclusion false.
To try to find such a counter example assign truth values so that the conclusion is false. A false conclusion would require that both $H$ and $R$ be true and at least one of $D$ and $S$ be false. With those values can one make all of the premises true? If both $H$ and $R$ are true then at least one of the two premises would be false, the one where either $D$ or $S$ was false. Hence there is no valuation of $H$, $R$, $D$ and $S$ that would make this invalid.
Alternatively, one could put this in a truth table generator. To do that conjoin the premises to form the sentence $(H\to D) \land (R\to S)$. Connect that sentence to the conclusion $(H \land R)\to (D \land S)$ using a conditional ($\to$) like so: $$((H\to D) \land (R\to S)) \to ((H \land R)\to (D \land S))$$
Any column reading false below that conditional connective would have valuations that make the conditional false. As it turns out, and as we can expect from the already existing proof, there aren't any:
Michael Rieppel. Truth Table Generator. https://mrieppel.net/prog/truthtable.html