I is true. Note that $m_{L_1} = \dfrac{-1}{m}$ and $m_{L_2} = \dfrac{-1}{p}$. According to the given figure, we have
$\begin{array}{rcl}
m_{L_1} & > & 0 \\
\dfrac{-1}{m} & > & 0 \\
m & < & 0
\end{array}$
and
$\begin{array}{rcl}
m_{L_2} & > & 0 \\
\dfrac{-1}{p} & > & 0 \\
p & < & 0
\end{array}$
Hence, we have
$\begin{array}{rcl}
m_{L_2} & > & m_{L_1} \\
\dfrac{-1}{m} & > & \dfrac{-1}{p} \\
\dfrac{1}{m} & > & \dfrac{1}{p} \\
p & > & m
\end{array}$
II is true. Note that the $x$-intercepts of $L_1$ and $L_2$ are $n$ and $q$ respectively. According to the given figure, we have
$\begin{array}{rcl}
n & > & q \\
\end{array}$
III is not true. Consider the intersection point of $L_1$ and $L_2$.
$\left\{\begin{array}{ll}
x + my = n & \ \ldots \unicode{x2460} \\
x + py = q & \ \ldots \unicode{x2461}
\end{array}\right.$
$\unicode{x2460} – \unicode{x2461}$, we have
$\begin{array}{rcl}
my – py & = & n -q \\
y & = & \dfrac{n-q}{m-p}
\end{array}$
According to the given figure, we have
$\begin{array}{rcl}
\dfrac{n-q}{m-p} & < & -1 \\
n - q & > & -m + p \\
m + n & > & p + q
\end{array}$
