Concentration of Hydrogen Ions
Since $\ce{NaOH}$ is a strong base, the pH can be directly calculated from the analytic concentration (1M):

$\ce{pOH} = -\log\ce{[OH-]} = -\log(1) = 0  $

$\ce{pH} = 14 - \ce{pOH} = 14 - 0 = 14$

For ammonia, you need to consider its reaction in water and the associated $\ce{K_b}$ value,

$\ce{NH3 + H2O -&gt; NH_4+ + OH-}$

The equilibrium concentrations for an unknown $x$ reacted amount (in concentration units) would be 

$\ce{[NH3]} = \ce{[NH3]_0} - x$ 

$\ce{[OH-]} = x$ 

$\ce{[NH4+]} = x$ 

$$\ce{K_b = \frac{[NH4+][OH-]}{[NH3]} = \frac{x^2}{[NH3]_0 - x}}$$

By plugging in the proper $\ce{K_b}$ value for ammonia and its analytical concentration ($\ce{[NH3]_0}$), you can calculate $x$, from which

$$\ce{pOH} = -\log(x)$$