Q is .10 m below P; mercury interfaces differ .05 m. A pressure balance gives $p_Q-p_P=\gamma_w(.10)+(\gamma_{Hg}-\gamma_w)(.05)$. Thus $9810[.10+(13.6-1)(.05)]=\mathbf{7161.3}$ Pa. The .08 m connecting-water segment cancels; using13.6 instead of 12.6 for the differential head would count water twice.
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