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There does not exist any function $p:F\mapsto p_F$ as in the question, to prove it by contradiction suppose such a function $p$ exists. In the following let $k+A=\{k+a;a\in A\}$ for any $k\in\mathbb{R}^2$, $A\subseteq\mathbb{R}^2$

To see why, consider the triangle $T_0$ with vertices $(-1,0),(1,0)$ and $(0,1)$. Let $r:(x,y)\mapsto(x,-y)$ be the reflection respect to the $x$ axis.

Now consider for each natural $n$ the triangle $T_n=r^n(T_0+10n)$, and let $p_n=p_{T_n}$ be given by $p_n=(x_n,y_n)$. Note that $d_H(T_n,T_{n+1})=10$, so $d(p_n,p_{n+1})\leq 10$. So $\sqrt{(x_{n+1}-x_n)^2+(y_{n+1}-y_n)^2}\leq 10$. This, of course, implies that $x_{n+1}-x_n\leq10$. Moreover, $(y_{n+1}-y_n)^2=(|y_{n+1}|+|y_n|)^2$, since $y_n$ and $y_{n+1}$ have opposite signs.

This will allow us to deduce by contradiction that $y_n\to 0$: if not, we would have some $\varepsilon>0$ and infinitely many $n$ such that $|y_n|>\varepsilon$, so $x_{n+1}-x_n<\sqrt{100-\varepsilon^2}$. This implies that for big enough $n$, we will have $x_n-x_0<10n-2$, which is impossible since $10n-2=d(T_n,T_0)$.

Similarly, let $S_0$ be the triangle with vertices $(-1,1),(1,1)$ and $(0,0)$, let $s$ be the reflection around the line $y=1$ and let $S_n=s^n(S_0+10n)$. Then letting $q_n=(z_n,w_n)=p_{S_n}$, we can prove that $w_n\to 1$ as before. So for big enough even $n$ we have $d_H(T_n,S_n)=\frac{\sqrt{2}}{2}$ and $d(p_n,q_n)\geq0.9$, the contradiction we were looking for.

There does not exist any function $p:F\mapsto p_F$ as in the question, to prove it by contradiction suppose such a function $p$ exists. In the following let $k+A=\{k+a;a\in A\}$ for any $k\in\mathbb{R}^2$, $A\subseteq\mathbb{R}^2$

To see why, consider the triangle $T_0$ with vertices $(-1,0),(1,0)$ and $(0,1)$. Let $r:(x,y)\mapsto(x,-y)$ be the reflection respect to the $x$ axis.

Now consider for each natural $n$ the triangle $T_n=r^n(T_0+10(n,0))$, and let $p_{T_n}=(x_n,y_n)$. Note that $d_H(T_n,T_{n+1})=10$, so $\sqrt{(x_{n+1}-x_n)^2+(y_{n+1}-y_n)^2}=d(p_{T_n},p_{T_{n+1}})\leq 10$. This, of course, implies that $x_{n+1}-x_n\leq10$. Moreover, $(y_{n+1}-y_n)^2=(|y_{n+1}|+|y_n|)^2$, since $y_n$ and $y_{n+1}$ have opposite signs.

This will allow us to deduce by contradiction that $y_n\to 0$: if not, we would have some $\varepsilon>0$ and infinitely many $n$ such that $|y_n|>\varepsilon$, so $x_{n+1}-x_n<\sqrt{100-\varepsilon^2}$. This implies that for big enough $n$, we will have $x_n-x_0<10n-2$, which is impossible since $10n-2=d(T_n,T_0)$.

Similarly, let $S_0$ be the triangle with vertices $(-1,1),(1,1)$ and $(0,0)$, let $s$ be the reflection around the line $y=1$ and let $S_n=s^n(S_0+10(n,0))$. Then letting $p_{S_n}=(z_n,w_n)$, we can prove that $w_n\to 1$ as before. So for big enough even $n$ we have $d_H(T_n,S_n)=\frac{\sqrt{2}}{2}$ and $d(p_n,q_n)\geq0.9$, the contradiction we were looking for.

It seems we cannot choose such a center if we want it to be preserved by isometries, that is, if for any isometry $i:\mathbb{R}^2\to\mathbb{R}^2$ and for any compact convex $F$ we want $p_{i(F)}=i(p_F)$ to be satisfied.

If that was the case, consider the triangle $T$ with vertices $P_1=(-1,0)$, $P_2=(1,0)$ and $P_3=(0,1)$. Then it is invariant respect to the reflection $r(x,y)=(-x,y)$, so we have $p_T$ has to be in the axis of the reflection, that is, $p_T=(0,y)$ for some $y$.

If $y>0$ consider the triangle $T_2=(5,0)-T:=\{(5,0)-q;q\in T\}$. Then $d_H(T,T_2)=10$, but $d(p_{T},p_{T_2})=d((0,y),(10,-y))>10$.

If $y=0$ we also obtain a contradiction, because letting $T_3=(0,\frac{1}{2})-T$, we have $d_H(T,T_3)=\frac{\sqrt{2}}{2}$ and $d(p_T,p_{T_3})=1$.

There does not exist any function $p:F\mapsto p_F$ as in the question, to prove it by contradiction suppose such a function $p$ exists. In the following let $k+A=\{k+a;a\in A\}$ for any $k\in\mathbb{R}^2$, $A\subseteq\mathbb{R}^2$

To see why, consider the triangle $T_0$ with vertices $(-1,0),(1,0)$ and $(0,1)$. Let $r:(x,y)\mapsto(x,-y)$ be the reflection respect to the $x$ axis.

Now consider for each natural $n$ the triangle $T_n=r^n(T_0+10n)$, and let $p_n=p_{T_n}$ be given by $p_n=(x_n,y_n)$. Note that $d_H(T_n,T_{n+1})=10$, so $d(p_n,p_{n+1})\leq 10$. So $\sqrt{(x_{n+1}-x_n)^2+(y_{n+1}-y_n)^2}\leq 10$. This, of course, implies that $x_{n+1}-x_n\leq10$. Moreover, $(y_{n+1}-y_n)^2=(|y_{n+1}|+|y_n|)^2$, since $y_n$ and $y_{n+1}$ have opposite signs.

This will allow us to deduce by contradiction that $y_n\to 0$: if not, we would have some $\varepsilon>0$ and infinitely many $n$ such that $|y_n|>\varepsilon$, so $x_{n+1}-x_n<\sqrt{100-\varepsilon^2}$. This implies that for big enough $n$, we will have $x_n-x_0<10n-2$, which is impossible since $10n-2=d(T_n,T_0)$.

Similarly, let $S_0$ be the triangle with vertices $(-1,1),(1,1)$ and $(0,0)$, let $s$ be the reflection around the line $y=1$ and let $S_n=s^n(S_0+10n)$. Then letting $q_n=(z_n,w_n)=p_{S_n}$, we can prove that $w_n\to 1$ as before. So for big enough even $n$ we have $d_H(T_n,S_n)=\frac{\sqrt{2}}{2}$ and $d(p_n,q_n)\geq0.9$, the contradiction we were looking for.

It seems we cannot choose such a center if we want it to be preserved by isometries, that is, if for any isometry $i:\mathbb{R}^2\to\mathbb{R}^2$ we want the center $p_F$ of any compact convex $F$ to satisfy $p_{i(F)}=i(p_F)$.

This is because if the center satisfied that, as the triangle $T$ with vertices $P_1=(-1,0)$, $P_2=(1,0)$ and $P_3=(0,1)$ is invariant respect to the reflection respect to the $y$-axis, we have $p_T=(0,y)$ for some $y$. If $y=0$ then we run into the counterexample mentioned in the question. If $y>0$ then consider the triangle $T_2=(5,0)-T:=\{(5,0)-q;q\in T\}$. Then $d_H(T,T_2)=10$, but $d(p_{T},p_{T_2})=d((0,y),(10,-y))>10$.

It seems we cannot choose such a center if we want it to be preserved by isometries, that is, if for any isometry $i:\mathbb{R}^2\to\mathbb{R}^2$ and for any compact convex $F$ we want $p_{i(F)}=i(p_F)$ to be satisfied.

If that was the case, consider the triangle $T$ with vertices $P_1=(-1,0)$, $P_2=(1,0)$ and $P_3=(0,1)$. Then it is invariant respect to the reflection $r(x,y)=(-x,y)$, so we have $p_T$ has to be in the axis of the reflection, that is, $p_T=(0,y)$ for some $y$. 

If $y>0$ consider the triangle $T_2=(5,0)-T:=\{(5,0)-q;q\in T\}$. Then $d_H(T,T_2)=10$, but $d(p_{T},p_{T_2})=d((0,y),(10,-y))>10$.

If $y=0$ we also obtain a contradiction, because letting $T_3=(0,\frac{1}{2})-T$, we have $d_H(T,T_3)=\frac{\sqrt{2}}{2}$ and $d(p_T,p_{T_3})=1$.