Answer
See the detailed answer below.
Work Step by Step
We need to find the potential due to one horizontal segment then multiply the result by 2, and then find the potential due to the semicircle segment.
$\color{blue}{ Horizontal\;segments:}$ See the first figure below:
We know that the potential at any point is the superposition of the potentials due to all charges.
So the potential at point $\rm p$ due to the $dq$ is given by
$$dV_{\rm p}=\dfrac{1}{4\pi \epsilon_0} \dfrac{dq}{r-x }\tag 1$$
Let's assume that the linear charge density of this rod is $\lambda$ which is given by
$$\lambda=\dfrac{Q}{L}=\dfrac{dq}{dx}$$
where $dq$ is the charge of the small black segment in the figure below and $dx$ is its length.
Thus,
$$dq=\dfrac{Qdx}{L}\tag 2$$
So the net electric potential is given by
$$V_{\rm p}=\dfrac{1}{4\pi \epsilon_0} \int_{-L/2}^{L/2} \dfrac{1}{ r-x }dq$$
Plug $dx$ from (1),
$$V_{\rm p}=\dfrac{1}{4\pi \epsilon_0} \int_{-L/2}^{L/2} \dfrac{1}{ r-x }\dfrac{Qdx}{L}$$
$$V_{\rm p}=\dfrac{Q}{(4\pi \epsilon_0)L} \int_{-L/2}^{L/2} \dfrac{1}{ r-x }dx$$
$$V_{\rm p}=\dfrac{Q}{(4\pi \epsilon_0)L}[-\ln(r-x)] \bigg|_{-L/2}^{L/2} $$
$$V_{\rm p}=\dfrac{-Q}{(4\pi \epsilon_0)L}\; \ln\left[\dfrac{\left(r-\dfrac{L}{2}\right)}{\left(r-\dfrac{-L}{2}\right)}\right] $$
$$V_{\rm p}=\dfrac{-Q}{(4\pi \epsilon_0)L}\; \ln\left[\dfrac{\left(r-\dfrac{L}{2}\right)}{\left(r+\dfrac{ L}{2}\right)}\right] $$
$$V_{\rm p}=\dfrac{ Q}{(4\pi \epsilon_0)L}\; \ln\left[\dfrac{\left(r+\dfrac{ L}{2}\right)}{\left(r-\dfrac{L}{2}\right)}\right] $$
Hence, the net potential due to the two segments at point $\rm p$ is given by
$$ (V_{\rm p})_{\rm Horizontal}=\dfrac{ 2Q}{(4\pi \epsilon_0)L}\; \ln\left[\dfrac{\left(r+\dfrac{ L}{2}\right)}{\left(r-\dfrac{L}{2}\right)}\right] $$
where here $r$, which is the distance from the center of the horizontal segment to point $\rm p$, $r=2R$, and $L$, which is the length of the horizontal segment, $L=2R$.
$$ (V_{\rm p})_{\rm Horizontal}=\dfrac{ 2Q}{(4\pi \epsilon_0)(2R)}\; \ln\left[\dfrac{\left(2R+\dfrac{ 2R}{2}\right)}{\left(2R-\dfrac{2R}{2}\right)}\right] $$
$$ (V_{\rm p})_{\rm Horizontal}=\dfrac{ 2Q}{(4\pi \epsilon_0)(2R)}\; \ln(3) $$
where $\lambda=Q/L=Q/(2R)$
$$ (V_{\rm p})_{\rm Horizontal}=\dfrac{ 2\lambda\ln(3)}{(4\pi \epsilon_0)}\tag A $$
$\color{blue}{ Semicircle\;segment:}$ See the second figure below:
We know that the potential at any point is the superposition of the potentials due to all charges. So the potential at point $\rm p$ due to the $dq$, in the given figure, is given by
$$dV=\dfrac{1}{(4\pi \epsilon_0)}\dfrac{dq}{R} \tag 1$$
Let's assume that the linear charge density of this rod is $\lambda$ which is given by $$\lambda=\dfrac{Q}{L}=\dfrac{dq}{ds}$$ where $dq$ is the charge of the small black segment in the figure below and $ds$ is its length.
Thus, $$dq=\dfrac{Qds}{L} $$
where $ds=Rd\theta$
$$dq=\dfrac{QRd\theta}{L} \tag 2$$
Plug $dq$ from (2) into (1),
$$dV=\dfrac{1}{(4\pi \epsilon_0)}\dfrac{1}{R}\dfrac{QRd\theta}{L}$$ where the length of the rod is now given by $L=\pi R$,
$$dV=\dfrac{1}{(4\pi \epsilon_0)} \dfrac{Q d\theta}{\pi R}$$
$$dV=\dfrac{1}{(4\pi \epsilon_0)} \dfrac{Q}{\pi R} d\theta$$
So the net electric potential at point $\rm p$ is given by
$$V_{\rm p}=\int dV=\dfrac{1}{(4\pi \epsilon_0)} \dfrac{Q}{\pi R}\int_{0}^{\pi} d\theta $$ $$V_{\rm p} =\dfrac{1}{(4\pi \epsilon_0)} \dfrac{Q}{\pi R}\theta\bigg|_0^\pi$$
$$V_{\rm p} =\dfrac{1}{(4\pi \epsilon_0)} \dfrac{Q}{\pi R} [\pi-0]$$
where $\lambda=Q/L=Q/(\pi R)$
$$ (V_{\rm p})_{\rm Semicircle} =\dfrac{\pi \lambda}{(4\pi \epsilon_0)} \tag B $$
Therefore, the net electric potential at point $\rm p$ which is the center of the semicircle is given by
$$(V_{\rm p})_{\rm net}=(V_{\rm p})_{\rm Semicircle} +(V_{\rm p})_{\rm Horizontal} $$
Plug from (A) and (B),
$$(V_{\rm p})_{\rm net}=\dfrac{\pi \lambda}{(4\pi \epsilon_0)} +\dfrac{ 2\lambda\ln(3)}{(4\pi \epsilon_0)}$$
$$\boxed{(V_{\rm p})_{\rm net}=\dfrac{\lambda}{(4\pi \epsilon_0)} \left[\pi + 2 \ln(3) \right]}$$
