Titration of multi-basic acid
$
\begin{align}
(n_{\ce{NaOH}})_i= \pu{0.2 mmol}\\
(n_{\ce{HCl}})_i= \pu{0.1 mmol}\\
(n_{\ce{NaH2PO4}})_i= \pu{0.1 mmol}\\
(n_{\ce{Na3PO4}})_i= \pu{0.05 mmol}
\end{align} 
$
At first the $\ce{NaOH}$ will react with $\ce{HCl}$ as per the following reaction:
$$
\begin{align}
\begin{array}{ccccc}
\ce{NaOH} &amp; + &amp; \ce{HCl} &amp; \ce{-&gt;} &amp; \ce{NaCl} &amp; + &amp; \ce{H2O} \\
\pu{0.2 mmol} &amp; &amp; \pu{0.1 mmol} &amp; &amp; \pu{0 mmol} &amp; &amp; \\
\pu{0.1 mmol} &amp; &amp; \pu{0 mmol} &amp; &amp; \pu{0.1 mmol} &amp; &amp; \\
\end{array}
\end{align}
$$
Now, the remaining $\ce{NaOH}$ will react with $\ce{NaH2PO4}$ as per following:
$$
\begin{align}
\begin{array}{ccccc}
\ce{NaOH} &amp; + &amp; \ce{NaH2PO4} &amp; \ce{-&gt;} &amp; \ce{Na2HPO4} &amp; + &amp; \ce{H2O} \\
\pu{0.1 mmol} &amp; &amp; \pu{0.1 mmol} &amp; &amp; \pu{0 mmol} &amp; &amp; \\
\pu{0 mmol} &amp; &amp; \pu{0 mmol} &amp; &amp; \pu{0.1 mmol} &amp; &amp; \\
\end{array}
\end{align}
$$
Now the resulting solution is  a buffer of $\ce{HPO4^2-}$ and $\ce{PO4^3-}$. So the Henderson equation can be used.
$
[\ce{PO4^3-}]=\frac{0.05}{45} \pu{M}\\
[\ce{HPO4^2-}]=\frac{0.1}{45} \pu{M}\\
$
$$
\begin{align}
\begin{array}{ccc}
\mathrm{pH} &amp; = &amp; \mathrm{pK_{a_3}} + \log(\frac{[\ce{PO4^3-}]}{[\ce{HPO4^2-}]}) \\
\mathrm{pH} &amp; = &amp; 12.35 + \log(\frac{0.05}{0.1})\\
\mathrm{pH} &amp; = &amp; 12.35 + \log(\frac{1}{2})\\
\mathrm{pH} &amp; = &amp; 12.04 \\
\end{array}
\end{align}
$$