Titration of H3PO4 with NaOH
The answer to your question does not need a single calculation.

You have got your 1 molar phosphoric acid that will get deprotonized by the 1 molar sodium hydroxide solution that is added. As both solutions have got the same concentration, you will reach the first inflection point after 100 mL and the second inflection point after 200 mL.

Now what is between those two volumes, at your desired volume of 150 mL?
A simple buffer system that can be described using the Henderson–Hasselbalch equation:
$$\mathrm{pH}=\mathrm pK_\mathrm a-\log_{10}\frac{\ce{c(H2A- )}}{\ce{c(HA^2- )}}$$

As both concentrations $\ce{c(H2A- )}$ and $\ce{c(HA^2- )}$ are equivalent, the logarithmic expression gets zero and your final pH equals the second $\mathrm pK_\mathrm a$ value of phosphoric acid, which is 7.21 (according to the German Wikipedia).

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