$p\Rightarrow \left(~\mathrm{q}\Rightarrow p\right)\equiv \sim p\vee \left(q\vee p\right)$
$\equiv \left(\sim p\vee p\right)\vee \left(q\right)\equiv t\vee \left(q\right)\equiv t$
Hence, the given statement $p\Rightarrow \left(~q\Rightarrow p\right)$ is a tautology
Now, $p\Rightarrow \left(p\Rightarrow q\right)$ is false when $p$ is true and $q$ is false

Hence, not a tautology
$p\Rightarrow \left(q\vee p\right)\equiv ~p\vee \left(p\vee q\right)\equiv \left(~p\vee p\right)\vee q$
$t\vee q\equiv t$
Hence, $p\Rightarrow \left(q\vee p\right)$ is a tautology.
Now, $p\Rightarrow \left(p\wedge q\right)$ is false when $p$ is true and $q$ is false

Hence, not a tautology
Also, $p\Rightarrow \left(p\iff q\right)$ is false when $p$ is true and $q$ is false

Hence, not a tautology