How to calculate the pH of a propanoate buffer after addition of sodium hydroxide?
The addition of the NaOH will "use up" 0.01 mol of propanoic acid and "create" 0.01 mol of propionate. To make things easier, just work in molarity. Thus 0.02 M NaOH.

The pH of the final solution can be found by adjusting the equation you used in part b,

$pH = -log \left( 1.26 \times 10^{-5}\frac{0.1-0.02}{0.05+0.02} \right)=4.84$

The result agrees with our expectations. Since the solution is a buffer, the pH is not expected to drastically change with the addition of a strong base. It is buffered.