Solution:
(Ⅰ) From the given conditions, we have $2pm=1$, which implies $m= \frac {1}{2p}$.
According to the definition of a parabola, we get $|PF|=m-(- \frac {p}{2})= \frac {1}{2p}+ \frac {p}{2}$.
Given that $\frac {1}{2p}+ \frac {p}{2}= \frac {5}{4}$, solving this equation yields $p= \frac {1}{2}$ or $p=2$ (discard this solution).
Therefore, the equation of $C$ is $y^{2}=x$.

(Ⅱ) Proof 1: Assume the slope of line $PA$ is $k$ (clearly $k\neq 0$), then the equation of line $PA$ is $y-1=k(x-1)$, thus $y=kx+1-k$.
By solving $\begin{cases} y=kx+1-k \\ y^{2}=x \end{cases}$ and eliminating $y$, we get $k^{2}x^{2}+[2k(1-k)-1]x+(1-k)^{2}=0$.
Let $A(x_{1},y_{1})$, by Vieta's formulas, we have $1\times x_{1}= \frac {(1-k)^{2}}{k^{2}}$, i.e., $x_{1}= \frac {(1-k)^{2}}{k^{2}}$. $y_{1}=kx_{1}+1-k=k\cdot \frac {(1-k)^{2}}{k^{2}}+1-k=-1+ \frac {1}{k}$. Thus, $A( \frac {(1-k)^{2}}{k^{2}},-1+ \frac {1}{k})$.
Given that the slope of line $PB$ is $\frac {1}{k}$.
Similarly, we get $B( \frac {(1- \frac {1}{k})^{2}}{( \frac {1}{k})^{2}},-1+ \frac {1}{ \frac {1}{k}})$, i.e., $B((k^{2}-1)^{2},k-1)$.
If the slope of line $l$ does not exist, then $\frac {(1-k)^{2}}{k^{2}}=(k-1)^{2}$. Solving this yields $k=1$ or $k=-1$.
When $k=1$, the slopes of lines $PA$ and $PB$ are both $1$, points $A$ and $B$ coincide, which contradicts the given conditions;
When $k=-1$, the slopes of lines $PA$ and $PB$ are both $-1$, points $A$ and $B$ coincide, which contradicts the given conditions.
Therefore, the slope of line $l$ must exist.
The equation of line $l$ is $y-(k-1)= \frac {k}{(k-1)^{2}}[x-(k-1)^{2}]$, i.e., $y= \frac {k}{(k-1)^{2}}x-1$.
Thus, line $l$ passes through the fixed point $(0,-1)$.

Proof 2: From (1), we have $P(1,1)$.
If the slope of $l$ does not exist, then $l$ is perpendicular to the $x$-axis.
Let $A(x_{1},y_{1})$, then $B(x_{1},-y_{1})$, $y_{1}^{2}=x_{1}$.
Then $k_{PA}k_{PB}= \frac {y_{1}-1}{x_{1}-1}\cdot \frac {-y_{1}-1}{x_{1}-1}= \frac {1-y_{1}^{2}}{(x_{1}-1)^{2}}= \frac {1-x_{1}}{(x_{1}-1)^{2}}= \frac {1}{1-x_{1}}$.
$(x_{1}-1\neq 0$, otherwise, $x_{1}=1$, then $A(1,1)$, or $B(1,1)$, line $l$ passes through point $P$, which contradicts the given conditions)
Given that $\frac {1}{1-x_{1}}=1$, so $x_{1}=0$. At this point, points $A$ and $B$ coincide, which contradicts the given conditions.
Therefore, the slope of $l$ must exist.
Assume the slope of $l$ is $k$ (clearly $k\neq 0$), let $l$: $y=kx+t$,
Since line $l$ does not pass through point $P(1,1)$, we have $k+t\neq 1$.
By solving $\begin{cases} y^{2}=x \\ y=kx+t \end{cases}$ and eliminating $y$, we get $k^{2}x^{2}+(2kt-1)x+t^{2}=0$.
By the discriminant $\triangle =1-4kt > 0$, we get $kt 0$.
Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then $y_{1}+y_{2}=n①$, $y_{1}y_{2}=-t②$
Then $k_{PA}k_{PB}= \frac {y_{1}-1}{x_{1}-1}\cdot \frac {y_{2}-1}{x_{2}-1}= \frac {y_{1}-1}{y_{1}^{2}-1}\cdot \frac {y_{2}-1}{y_{2}^{2}-1}= \frac {1}{y_{1}y_{2}+(y_{1}+y_{2})+1}$.
Given that $y_{1}y_{2}+(y_{1}+y_{2})+1=1$, i.e., $y_{1}y_{2}+(y_{1}+y_{2})=0③$
Substituting ①② into ③ yields $-t+n=0$, i.e., $t=n$.
Therefore, $l$: $x=n(y+1)$. Clearly, $l$ passes through the fixed point $(0,-1)$.

Thus, for all proofs, the conclusion is that line $l$ passes through the fixed point $\boxed{(0,-1)}$.