Let $p$ be the $x$-coordinate of the intersection of $l_P$ and $y=1$.
Let $q$ be the $x$-coordinate of the intersection of $l_Q$ and $y=1$.
If $p<1$ then $p$ is uniformly distributed in $[0,1]$, because the probability that $p<x$ is proportional to the area of the triangle with vertices $(0,0)$, $(0,1)$ and $(x,1)$, which is proportional to $x$.
Similarly, if $q>0$ then $q$ is uniformly distributed in $[0,1]$.
For the lines not to intersect inside the square, first we require that $p<1$, which has probability $\frac12$. Then we require that $q>0$, which has probability $\frac12$. Finally we require that $p<q$, which (given that the first two requirements have been met) has probability $\frac12$ (since $p$ and $q$ are each uniformly distributed in $[0,1]$). So the probability that the lines do not intersect inside the square, is $\left(\frac12\right)^3=\frac18$.
Therefore the probability that the lines intersect inside the square is $\frac78$.