Does NaOH have a higher OH- concentration than NH3?
Given that the solutions have equal concentrations, a solution of $\ce{NaOH}$ (a strong base) will be more basic than one of $\ce{NH3}$ (a weak base), since $\ce{NH3}$ only dissociates partially:

$$\ce{NH3 + H2O &lt;&lt;=&gt; NH4+ + OH-}$$

Loosely speaking, the pH of a solution is defined to be

$$\mathrm{pH} = -\log_{10}[\ce{H+}]$$

and from the ionic product of water, we know that $[\ce{H+}][\ce{OH-}]$ is always a constant (equal to $10^{-14}\ \mathrm{M^2}$ at $25\ ^\circ\mathrm{C}$).

Therefore, a higher pH implies a lower concentration of $\ce{H+}$, and therefore a higher concentration of $\ce{OH-}$. This means that the truth of option 1 implies the truth of option 2, and vice versa, i.e. option 1 is logically equivalent to option 2.

My first suspicion would therefore be a typo in your book. In any case, your reasoning is correct.