Finding is the concentration of H⁺ in a benzoate buffer solution after adding HCl
Initially, it's true that you have the following equilibrium between sodium benzoate and benzoic acid:
$$\ce{C6H5COOH + H2O &lt;=&gt; C6H5COO- + H3O+_\mathrm{(aq)}}$$
When you add  $\ce{HCl}$ to your buffer solution, sodium benzoate reacts with the acid, and the aforementioned equilibrium is displaced to the left:
$$\ce{C6H5COO- + H3O+_\mathrm{(aq)}  &lt;=&gt;  C6H5COOH + H2O }$$
Now, the equilibrium constant of this reaction is:
$$K=\frac{1}{K_ a}=0.16 \times 10^5&gt;10^3$$
The reaction is quantitative (total). So, we can rewrite it:
$$\ce{C6H5COO- + H3O+_\mathrm{(aq)} -&gt;  C6H5COOH + H2O }$$
Now, we have to construct an ICE table:
Assuming that we didn't dilute the solution when we added $\ce{HCl}$ (this point is not clear in your question): $$\ce{[C6H5COOH]_0} = \pu{0.12 M}$$
$$\ce{[C6H5COO^-]_0} =  \pu{0.21 M}$$
$$\ce{[H^+]_0} = (\frac{0.050}{1.3} + 10^{-4.4}) \mathrm{M}\approx 0.0385 \mathrm{M}$$
As you can clearly seen $\ce{[H^+]_0}$ is the lowest, and so ion hydronium is the limiting reactant.
At the end of the reaction, we have:
$$\ce{[C6H5COO^-]_\mathrm{(eq)}} = (0.21-0.0385) = \pu{0.1715 M}$$
$$\ce{[C6H5COOH]_\mathrm{(eq)}} = (0.12+0.0385) = \pu{0.1585 M}$$
$$\ce{[H^+]_\mathrm{(eq)}} = \epsilon$$
Substituting all these numbers in the expression of $K$
$$K = \frac{ 0.1585}{0.1715\epsilon}=0.16 \times 10^5$$
By solving the above equation, we find:
$$\ce{[H^+]} = \epsilon = \pu{5.82 \times 10^{-5} M}$$
$$\mathrm{pH} = 4.23$$