Should the centripetal force of Earth's orbit around the Sun affect a pendulum on Earth? Lets use approximate earth angular velocity around the sun with $\omega\approx2*10^{-7}\frac{r}{s}$ and earth translation orbit radius with $r\approx1.5*10^{11}m$ we can approximate centripetal acceleration with $|A_{c}|=\omega^2*r\approx0.006\frac{m}{s^2}$ and $A_c$ is pointing to the center of the orbit. Is every object on earth affected by that acceleration with an absolute value of $|A_c|=0.006\frac{m}{s^2}$? If so lets consider we have a static pendulum without any other force applied to it, at a specific time and place such that $A_c$ is completely perpendicular to it, we should see the pendulum a little displaced to one side and 12 hours later a little bit displaced to the opposite side? (since earth makes half turn and now pendulum is oriented to the opposite side). I know since also the moon orbits the earth, we can repeat the same argument, but all things consider it's correct to try to predict with a immobile pendulum a little deviation of its center?
 A: Yes and no. ;) A body in orbit is in freefall and experiences no weight due to the gravity of the central body which it orbits. A pendulum in the International Space Station wouldn't work because it's weightless. 
To a first approximation, the Earth and everything on it is weightless relative to the Sun. But that's only approximately true because the Earth is an extended body, not a point. It's more accurate to say that the centre of the Earth is in freefall and all other points on the Earth feel a slight force from the Sun because their velocity isn't exactly the proper velocity for a point at that orbital distance. That slight force is called a tidal force because it drives the ocean tides.
But even that's not quite true because the Earth has a rather large Moon. The Earth and Moon orbit around their common centre of mass, their barycentre, which is located (on average) about 1700 km below the surface of the Earth, or about 4670 km from the centre of the Earth. 
So the Earth-Moon barycentre is in freefall around the Sun, and anything on Earth not located at that barycentre will experience a tidal force from the Sun. It will also experience a tidal force from the Moon, and in fact the tidal force from the Moon is larger than that from the Sun because tidal forces diminish in accordance with an inverse cube law, rather than the inverse square law of direct gravitational forces.
On the Earth's surface, these tidal forces are quite small compared to the gravitational force of the Earth. Relative to $g$, the mean gravitational acceleration at the surface of the Earth, the lunar tidal acceleration is approximately $1.12×10^{-7}g$ and the solar tidal acceleration is approximately $5.14×10^{-8}g$.
It is possible to detect those tidal accelerations with a pendulum, but the pendulum needs to be built with very high precision and very low friction. And you need a very good clock to measure the tidal deviations in the pendulum's period with sufficient precision.
"Time hacker" Tom Van Baak has written an excellent series of articles on this topic. Tom has performed many tests using atomic clocks and pendulums, and his articles are copiously illustrated with graphs. They also contain many equations, but (generally) don't require mathematical knowledge beyond high school level. You can find links to these articles here:  
Precision Pendulum Clocks, Gravity and Tides
However, your question asks about angular deviations of a static pendulum. The tidal forces will cause such deviations, but they are very hard to measure accurately. Measuring the period of a swinging pendulum would be a lot easier.
A: There is no displacement. As you said, every object on earth is affected by the centripetal acceleration. That includes the mass of the pendulum, but also the wire, the air around and the ceiling where it is attached. 
A similar and easier way to test that reasoning is placing a weight about 3 kg on a kitchen scale, with a precision of 1g. 
If the centripetal acceleration of the sun had any effect, the scale display would show a difference of more than 1g when doing the test at midnight or at noon. (The test can be done in a day of new moon to add moon and sun effect).
But as the pendulum, the weight on the scale and the scale itself are orbiting the sun, and are subject to the same centripetal acceleration, and no difference can be measured.
A: Yes, it would, with amplitude depending on location. 
To see this empirically, note that solar tides exit:  the combination of the centripetal/centrifugal force and the gravity gradient of the Sun result in differential forces on the ocean, and by analogy on your pendulum. 
Do tides involve the centripetal/centrifugal forces?  Yes, it does. To see that, consider how you get a tidal bulge on the side of the Earth away from the Sun. 
A: None of this the addresses the sun and solar systems free fall around the center of the galaxy or our galaxy’s movement through the universe. Impossible to measure with a pendulum and clock?
