Nitric oxide is a tough molecule to represent as a Lewis structure. However, you have made one common mistake in your structures. I also want to clear up a misconception about resonance that's present in your post. Finally, I will need to introduce a different bonding picture of $\ce{NO}$ to answer your question about which orbital the unpaired electron is in.
The Lewis Structure of $\ce{NO}$
The mistake you made in drawing those Lewis Structures is that you used all of the electrons in the nitrogen and oxygen atoms and not just the valence or outer-shell electrons.
A Nitrogen atom has 7 total electrons in the following configuration:
$$1s^2 2s^2 2p_x^1 2p_y^1 2p_z^1$$
Of these seven electrons, only the $5$ in the second shell (energy level) are available for bonding. These are the valence electrons. The $1s$ electrons are core electrons. They are not available for bonding because their probability density function maxes at less than the atomic radius.
Similarly, an oxygen atom can only use the six electrons in its second shell for bonding: $$2s^2 2p_x^2 2p_y^1 2p_z^1$$
The bonding in $\ce{NO}$ thus only has 11 electrons and not 13 (like you have). With two fewer electrons, it's easy to place the unpaired electron in a way that does not violate the octet rule.
$$\begin{aligned}
&.. \ \ \ .. &&\ &&&\ ..\ \ \ ..\ \\
\ce{.} &\ce{N=O:} &&\ce{<->} &&&\ce{:N=O.}
\end{aligned}$$
Resonance
The most common misconception related to resonance is unfortunately suggested by the name of the phenomenon. The two resonance structures drawn do not exist as separate structures inter-converting. They are attempts to represent a compound using Lewis structures that has bonding that is not localized. The true structure of nitric oxide is some hybrid of these resonance contributors.
Where is the unpaired electron?
To answer this question, we need molecular orbital theory. A molecular orbital diagram describes the bonding in terms of constructive and destructive overlap of atomic orbitals. A MO diagram dispenses with the need for resonance - all electrons are clearly shown in their orbitals: $\sigma$ or $\pi$, bonding or anti-bonding. Since this answer is running long already, I won't go into how to construct such a diagram.
Here is the MO diagram for nitric oxide:
The unpaired electron (highlighted in blue)is in a $\pi$ anti-bonding orbital that is formed from the destructive overlap of out of phase $p$ orbitals, one from nitrogen and one from oxygen.
