The statement $\left(p\wedge \left(~q\right)\right)\Rightarrow \left(p\Rightarrow \left(~q\right)\right)$ is

Using logical equivalence condition $\left(p\to q\equiv ~p\vee q\right)$, given statement can be expressed as, 

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

$\equiv \left(~\left(p\wedge ~q\right)\right)\vee \left(~p\vee ~q\right).......\left(1\right)$

We know De Morgan's law of logical equivalence is,

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

Applying De Morgan's law in equation(1) we get,

$\equiv \left(~p\vee q\right)\vee \left(~p\vee ~q\right)$   $\left[\because ~\left(~p\right)\equiv p\right]$

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

$\equiv ~p\vee t\equiv t$ (tautology)