$(P\implies Q) \implies [(R ∨ P)\implies (R ∨ Q)]$ is a tautology

HINT: $(P\to Q)\to\lnot P\lor Q\lor R\equiv(P\to Q)\to((P\to Q)\lor R)$.

You can take benefit of the fact that :

$(A \implies B)$ is equivalent to $(\lnot A \lor B)$

so that, e.g

$(R \lor P)$ will be $(\lnot R \implies P)$.

Now, if you rewrite your formula in this way, you will get :

$$(P \implies Q) \implies [ (\lnot R \implies P) \implies (\lnot R \implies Q)]$$

and this is a version of Hypothetical Syllogism.

P   Q   R   (P-->Q)   (R V P)  (R V Q)  [(R V P) --> (R V Q)] {(P-->Q) ->[(R V P)-->(R V Q)]}

T   T   T     T          T        T               T                     T

T   T   F     T          T        T               T                     T

T   F   T     F          T        T               T                     T

T   F   F     F          T        F               F                     T

F   T   T     T          T        T               T                     T

F   T   F     T          F        T               T                     T

F   F   T     T          T        T               T                     T

F   F   F     T          F        F               T                     T

The last column as all the T's which means your statement is a tautology. This is probably easier to do than use equivalent statements.