Điều kiện: \(y < 4\).

\(\log x + {x^2} + xy \ge \log \left( {4 - y} \right) + 4x \Leftrightarrow \log x + {x^2} \ge \log \left( {4 - y} \right) + 4x - xy\)

\( \Leftrightarrow 2\log x + {x^2} \ge \log \left( {4 - y} \right) + \log x + x\left( {4 - y} \right) \Leftrightarrow \log {x^2} + {x^2} \ge \log \left( {4 - y} \right)x + x\left( {4 - y} \right)\,\,(1)\)

Xét hàm số \({\rm{f}}\left( {\rm{t}} \right) = \log {\rm{t}} + {\rm{t}}\,\,\forall {\rm{t}} \in \left( {0\,;\,\, + \infty } \right) \Rightarrow {\rm{f'}}\left( {\rm{t}} \right) = \frac{1}{{{\rm{t}} \cdot \ln 10}} + 1 > 0\,\,\forall {\rm{t}} \in \left( {0\,;\,\, + \infty } \right)\)

\((1) \Rightarrow f\left( {{x^2}} \right) = f\left( {\left( {4 - y} \right)x} \right) \Leftrightarrow x = 4 - y \Leftrightarrow x + y = 4\).

\(P = 8x + 16y + \frac{1}{x} + \frac{{147}}{y} = 4x + \frac{1}{x} + 12y + \frac{{147}}{y} + 4\left( {x + y} \right)\)

\( \Rightarrow P \ge 2 \cdot \sqrt {4x \cdot \frac{1}{x}} + 2 \cdot \sqrt {12y \cdot \frac{{147}}{y}} + 4 \cdot 4 = 104\)

\( \Rightarrow {P_{\min }} = 104 \Leftrightarrow y = \frac{7}{2};x = \frac{1}{2}.{\rm{ }}\)Đáp án: 104.