Buffer Solution - Sodium Benzoate
In that equation, the acid and base are a conjugate acid/base pair.  We assume HCl quantitatively dissociates and protonates benzoates.  The concentration of the acid increases, and the base decreases. 

Empirically, the pH should go down in this case.  Without seeing your specific calculation, we can't comment where you are going wrong, but the proper flow should look something like:

$$\mathrm{pH=pK_a+Log_{10}(\frac{[base]-[Added H]}{[acid]+[AddedH]})}$$

$$\mathrm{pH=pK_a+Log_{10}(\frac{0.39M - 0.036M}{0.10M + 0.036M})}$$

$$\mathrm{pH=pK_a+Log_{10}(\frac{0.354M}{0.136M})}$$
From your calculated $\mathrm{pKa = 4.20}$, you should get there pretty quickly.  

Just applying the difference in the log values of the changed ratios, I get  $\mathrm{pH = 4.61}$