(I) According to the problem, the parabola $\Gamma :{{y}^{2}}=2px$ passes through point $A(4,4)$.

So, we substitute the coordinates of point $A$ into the equation: ${{4}^{2}}=2p\cdot 4 \Rightarrow p=2$.

Therefore, the equation of the parabola $\Gamma$ is: ${{y}^{2}}=4x$.

(II) From part (I), we know that the focus $F$ of the parabola $\Gamma$ has coordinates $F(1,0)$, and we also have point $A(4,4)$.

The slope of line $l$ is given by: $k=\\frac{4-0}{4-1}=\\frac{4}{3}$.

So, the equation of line $l$ is: $y=\\frac{4}{3}(x-1)$, which can be rewritten as $4x-3y-4=0$.

Substituting the equation of the parabola ${{y}^{2}}=4x$ into the equation of line $l$, we get the coordinates of point $B$ as $B(\\frac{1}{4},-1)$.

Hence, the length of segment $AB$ is given by: $\\left| AB \\right|=\\sqrt{{{(4-\\frac{1}{4})}^{2}}+{{(4+1)}^{2}}}=\\sqrt{\\frac{625}{16}}=\\boxed{\\frac{25}{4}}$.