Given that $L_1$ is decidable language. That means it has a TM  ($M_1$) decider.Now we consider a language, $L_p$;$$L_p = \left \{ \langle M \rangle \mid L(M) \text{ is reducible to } L(M_1) \right \}$$$L_p$ consists of all the encodings of the Turing machines whose language is reducible to $L_1$.What can we say about $L_p$, Or, $L_p$ is decidable or undecidable ?I understand if we assume to have a decider for $L_p$ and using this as component if we can construct a TM which decides a problem $P$. If $P$ is already known as undecidable then we can conclude that out assumed decider for $L_p$ does not exist.How to come up with this reduction idea for $L_p$ ?
Decidability of the language of all TMs whose languages reduce to a fixed decidable language
computability;turing machines;reductions;undecidability