How to find the concentration of an of ammonia solution by titration with nitric acid?
Yes, although your presentation is a bit confusing.
To be clear,


You determined the quantity of $\ce{HNO3}$ consumed during the titration 
$$0.151\ \mathrm{M} \times 26.2\ \mathrm{mL} = 3.956 \  \mathrm{mmol}$$
Then given this titration equation:
$$\ce{HNO3 + NH3 -&gt; NO3- + NH4+}$$
You deduced that the $25~\mathrm{ml}$ sample contained $3.956 \  \mathrm{mmol}$ of $\ce{NH3}$, hence a concentration of 
$$\frac{3.956\ \mathrm{mmol}}{25\ \mathrm{mL}}=0.1582\ \mathrm{M}$$
From which you calculated the amount of $\ce{NH3}$ in the $500\ \mathrm{mL}$ sample:
$$0.1582\ \mathrm{M} \times 500\ \mathrm{mL} = 79.12 \  \mathrm{mmol}$$
And finally since the original $20\ \mathrm{mL}$ sample also contains $79.12\ \mathrm{mmol}$ of $\ce{NH3}$, that gives you the original concentration:
$$\frac{79.12\ \mathrm{mmol}}{20\ \mathrm{mL}}=3.956\ \mathrm{M}$$


What I think you should have done in place of 4 and 5 is:

4'. Since the $25\ \mathrm{mL}$ sample is taken from a solution diluted from $20\ \mathrm{mL}$ to $500\ \mathrm{mL}$, the original concentration is:
$$0.1582 \ \mathrm{M} \times \frac{500\ \mathrm{mL}}{20\ \mathrm{mL}}=3.956\ \mathrm{M}$$

Which spares you one calculation's work and potential errors.