Using a metric on a metric space $(X,d)$ we can define Hausdorff metric $h$ on the set of all compact subsets of $(X,d)$,say $K(X)$. I am able to show that $h$ is indeed a metric. Suppose instead of metric $d$ on $X$ we had a partial metric $p$ on $X$. If we analogously define Hausdorff partial metric, say $h_p$, similar to the definition of $h$, will $h_p$ be a partial metric non $K(X)$?
Edit
Let $X\ne \emptyset$. A partial metric on $X$ is a function $p:X\times X \to \mathbb{R}^+$ satisfying
a)$p(x,y)=p(y,x)$ (symmetry)
b)$0\le p(x,x) =p(x,y)= p(y,y)$ then $x=y$ (equality)
c)$p(x,x)\le p(x,y)$ (small self distances)
d)$p(x,z)\le p(x,y)+p(y,z)-p(y,y)$ (triangle inequality)
for all $x,y,z\in X.$
Why partial metric is useful?
Some research articles state that, partial metric is useful in modeling partially defined information as comes in computer science.
Definition
$$h_p(A,B)=\sup_{a\in A}\inf_{b\in B} p(a,b)$$
$K(X)$: set of all compact subsets in the partial metric space $(X,p)$
Question
Is $h_p$ a partial metric on $K(X)$?
Not necessarily. Let $X$ be the first of the examples of partial metric space from the site linked by you, that is $X=\Bbb R^+$ with $p(x,y)=\max\{x\}$. Let $A=\{1,2,3\}$ and $B=\{1,3\}$ be elements of $K(X)$. Then $$h_p(A,A)= h_p(A,B)= h_p(B,B),$$ but $A\ne B$, which violates the equality of $h_p$.