We know,

$A\to B\equiv ~A\vee B$

Now solving all options one by one we get,

$\left(A\right) p\to \left(p\wedge \left(p\to q\right)\right)$

$\equiv \left(~p\right)\vee \left(p\wedge \left(~p\vee q\right)\right)$

$\equiv \left(~p\right)\vee \left(f\vee \left(p\wedge q\right)\right)$

$\equiv ~p \vee \left(p\wedge q\right)\equiv \left(~\mathrm{p}\vee \mathrm{p}\right)\wedge \left(~p\vee q\right)$

$\equiv ~p \vee q$

Which is not a tautology

$\left(B\right) \left(p\wedge q\right)\to \left(~p\to q\right)$

$\equiv ~\left(p\wedge q\right)\vee \left(p\vee q\right)$

$\equiv ~p\vee ~q\vee \left(p\vee q\right)$

$\equiv ~p\vee p\vee ~q\vee q$

$\equiv T\vee T\equiv T$

Tautology

$\left(C\right) \left(p\wedge \left(p\to q\right)\right)\to ~q$

$\equiv ~\left(p\wedge \left(~p\vee q\right)\right)\vee ~q$

$\equiv ~\left(p\wedge q\right)\vee ~q$

$\equiv ~p\vee ~q$

Which is not a tautology

$\left(D\right) p\vee \left(p\wedge q\right)\equiv p$

Which is not a tautology