EDIT: Given updated edit to question, it seems we want to prove:
$[P \rightarrow (Q\land R)] \iff [(P\rightarrow Q) \land (P\rightarrow R)]$
It's usually best to prove $\iff$ equivalence-statements by proving each direction of the double implication $(\;\Rightarrow\;$ AND $\;\Leftarrow\;)$, so you don't make an inadvertent mistake by using a rule of inference that might go only one direction.
So you want to prove each of the following two implications:
$[(P\rightarrow Q) \land (P\rightarrow R)] \implies [(P \rightarrow (Q\land R)] \tag{1}$
$[P\rightarrow (Q\land R)] \implies [(P \rightarrow Q) \land (P \rightarrow R)]\tag{2}$
Hints:
for both (1) and (2): Use the identity that
$(A \rightarrow B) \equiv (\lnot A \lor B)\quad$ for each of the inner implications ($\rightarrow$).
Then you can use distribution:
$[(A\lor B) \land C] \equiv [(A \land C) \lor (B \land C)];\quad[(A\land B) \lor C] \equiv [(A\lor C) \land (B \lor C)]$
And use DeMorgan's Laws;
$\lnot (A \land B) \equiv (\lnot A \lor \lnot B),\quad \text{and} \quad \lnot (A \lor B) \equiv (\lnot A \land \lnot B)$
Once you prove $(1)$, in this case you can prove $(2)$ by "undoing" what you did to prove $(1)$, starting with "undoing" the last step in the proof of $(1)$, and working backwards.
Note that each implication $(1)$ and $(2)$ is a tautology: both implications are true for all possible truth-value assignments to $P, Q, R$:
