Solution:
$(1)$ Given that $C_{1}(1,0)$, let point $D(x,y)$ be any point on the parabola $C_{1}$. Then $|DC_{1}|= \sqrt {(x-1)^{2}+y^{2}}= \sqrt {x^{2}-2(1-p)x+1} (x\geqslant 0)$.
Let $f(x)=x^{2}-2(1-p)x+1$, $x\in[0,+\infty)$, then $f(x)$ has its minimum value $1$ only when $x=0$.
If $0 < p < 1$, then $f(x)$ reaches its minimum value when $x=1-p$. Setting $1-p=0$ gives $p=1$, which is a contradiction;
If $p\geqslant 1$, then $f(x)$ reaches its minimum value $1$ when $x=0$, which meets the requirement.
Therefore, $\boxed{p\geqslant 1}$.
$(2)$ Let the slope of line $MA$ be $k$, and the slope of line $MB$ be $-k$, where $k\neq 0$.
The equation of line $MA$ is $y-y_{0}=k(x-x_{0})$. Substituting $x= \frac {y^{2}}{2p}$ and rearranging gives $ky^{2}-2py+2py_{0}-2pkx_{0}=0$.
Then $y_{A}+y_{0}= \frac {2p}{k}$, so $y_{A}= \frac {2p}{k}-y_{0}$.
Also, $y_{A}-y_{0}=k(x_{A}-x_{0})$, rearranging gives $x_{A}= \frac {2p}{k^{2}}- \frac {2y_{0}}{k}+x_{0}$.
Substituting $k$ with $-k$ gives $x_{B}= \frac {2p}{k^{2}}+ \frac {2y_{0}}{k}+x_{0}$, $y_{B}=- \frac {2p}{k}-y_{0}$.
Therefore, the slope of line $AB$, $K_{AB}= \frac {y_{B}-y_{A}}{x_{B}-x_{A}}= \frac {- \frac {2p}{k}-y_{0}-\left( \frac {2p}{k}-y_{0}\right)}{ \frac {2p}{k^{2}}+ \frac {2y_{0}}{k}+x_{0}-\left( \frac {2p}{k^{2}}- \frac {2y_{0}}{k}+x_{0}\right)}=- \frac {p}{y_{0}}$.
Thus, the slope of line $AB$ is a non-zero constant, $\boxed{- \frac {p}{y_{0}}}$.