Sum of products

Notice that if $x+y>xy$ ,then one of $x$ and $y$ must equal 1.

If $a+b=ab,(a-1)(b-1)=1,a=b=2$ ,and also $c+d=cd,c=d=2$ .

If $a+b<ab$ ,then $c+d>cd$ ,and that means $c=1$ .We get $a+b=d,ab=d+1,ab-a-b=1,(a-1)(b-1)=2,a=2,b=3,d=5$

If $a+b>ab$ ,then $a=1$ ,that means $c=1$ .According to the above equation,if $c=1$ then $a=2$ ,so it doesn't fit.

Hence,P=16 or 30.

$\displaystyle\begin{cases}a+b=cd \\ c+d=ab \end{cases}$ implies that $\displaystyle\begin{cases} \frac{a+b}{cd}=1\\ \frac{c+d}{ab}=1\end{cases}$ . This yields $\displaystyle \left( \frac{1}{a}+\frac{1}{b}\right)\left( \frac{1}{c}+\frac{1}{d}\right)=1$ .

Suppose $\displaystyle \frac{1}{a}+\frac{1}{b}=1$ . Then $\displaystyle a+b=ab$ which means that $\displaystyle (a-1)(b-1)=1$ . Since $a,b \in \mathbb{N}$ , $a-1=b-1=1$ , or equivalently, $a=b=2$ . The assumption $\displaystyle \frac{1}{a}+\frac{1}{b}=1$ also implies that $\displaystyle \frac{1}{c}+\frac{1}{d}=1$ , which also means that $c=d=2$ . Hence $\color{#3D99F6}{ (a,b,c,d)=(2,2,2,2)}$ is a solution.

If $\displaystyle \frac{1}{a}+\frac{1}{b}>1$ , then $a=1$ or $b=1$ . However, this is impossible as $c\le a\le b\le d$ implies that $a=c=1$ which means $\displaystyle \left( \frac{1}{a}+\frac{1}{b}\right)\left( \frac{1}{c}+\frac{1}{d}\right)=\left( 1+\frac{1}{b}\right)\left( 1+\frac{1}{d}\right)>1$ .

Therefore $\displaystyle \frac{1}{a}+\frac{1}{b}<1$ . It is clearly seen that $\displaystyle \frac{1}{c}+\frac{1}{d}>1$ which implies that $c=1$ . From the equations $\displaystyle\begin{cases}a+b=cd \\ c+d=ab \end{cases}$ , we now have $\displaystyle\begin{cases}a+b=d \\ 1+d=ab \end{cases}$ . By eliminating $d$ , we will obtain $a+b=ab-1$ which means that $(a-1)(b-1)=2$ . Since $a,b \in \mathbb{N}$ , $a=2$ and $b=3$ . Lastly, $d=a+b=5$ . Hence $\color{#3D99F6}{ (a,b,c,d)=(2,3,1,5)}$ is another solution.

Finally, $P=2^4+2(3)(1)(5)=\boxed{46}$ .