You can see that when $p=0, q=1, r=1$, the LHS is True but the RHS is False.
You can reduce the original equation as follows, which also should tell you how you can arrive at those values for the variables.
$$\begin{align}
&\Bigl [ (\lnot p \land q) \land \bigl[ q \to (p \to q) \bigr ] \Bigr ] \to \lnot r \\[1em]
\equiv& \Bigl [ \bar p \cdot q \cdot \bigl[ q \to (p \to q) \bigr ] \Bigr ] \to \bar r & \text{easier notation}\\[0.5em]
\equiv& \Bigl [ \bar p \cdot q \cdot \bigl[ \bar q + (p \to q) \bigr ] \Bigr ] \to \bar r & x \to y \equiv \bar x + y\\[0.5em]
\equiv& \Bigl [ \bar p \cdot q \cdot \bigl[ \bar q + (\bar p + q) \bigr ] \Bigr ] \to \bar r \\[0.5em]
\equiv& \Bigl [ \bar p \cdot q \cdot \bigl[ 1 \bigr ] \Bigr ] \to \bar r & \bar x + x \equiv 1\\[0.5em]
\equiv& \Bigl [ \bar p \cdot q\Bigr ] \to \bar r & x \cdot 1 \equiv x\\[0.5em]
\equiv& \overline {( \bar p \cdot q)} + \bar r \\[0.5em]
\equiv& ( p + \bar q) + \bar r & \overline{x\cdot y} \equiv (\bar x + \bar y), \quad \bar{\bar x} = x\\[0.5em]
\not \equiv& 1
\end{align}$$
