Official solution
Let \[\theta\] = tan-1\[\left( {{{12} \over 5}} \right)\]
\[ \Rightarrow \] tan\[\theta\] = \[{{{12} \over 5}}\]
\[ \Rightarrow \] \[{{2\tan {\theta \over 2}} \over {1 - {{\tan }^2}{\theta \over 2}}} = {{12} \over 5}\]
\[ \Rightarrow \tan {\theta \over 2} = {2 \over 3}\]
\[ \Rightarrow \sin {\theta \over 2} = {2 \over {\sqrt 3 }}\] and \[\cos {\theta \over 2} = {3 \over {\sqrt {13} }}\]
In \[\Delta\]CAP,
\[\tan {\theta \over 2} = {1 \over {AP}}\]
\[ \Rightarrow AP = {3 \over 2}\]
In \[\Delta\]APM,
\[\sin {\theta \over 2} = {{AM} \over {AP}},\cos {\theta \over 2} = {{PM} \over {AP}}\]
\[ \Rightarrow AM = {3 \over {\sqrt {13} }}\]
\[ \Rightarrow PM = {9 \over {2\sqrt {13} }}\]
\[ \therefore \] \[AB = {6 \over {\sqrt {13} }}\]
\[ \therefore \] Area of \[\Delta PAB = {1 \over 2} \times AB \times PM\]
\[ = {1 \over 2} \times {6 \over {\sqrt {13} }} \times {9 \over {2\sqrt {13} }} = {{27} \over {16}}\]
Now, \[\phi = 90^\circ - {\theta \over 2}\]
In \[\Delta\]CAM,
\[\cos \phi = {{CM} \over {CA}}\]
\[ \Rightarrow CM = 1.\cos \left( {{\pi \over 2} - {\theta \over 2}} \right)\]
\[ = 1.\sin {\theta \over 2} = {2 \over {\sqrt {13} }}\]
\[ \therefore \] Area of \[\Delta CAB = {1 \over 2} \times AB \times CM\]
\[ = {1 \over 2} \times {6 \over {\sqrt {13} }} \times {2 \over {\sqrt {13} }} = {6 \over {13}}\]
\[ \therefore \] \[{{Area\,of\,\Delta PAB} \over {Area\,of\,\Delta CAB}} = {{27/26} \over {6/13}} = {9 \over 4}\]
Therefore, the correct answer is (2).