In a sequence $\left\{a_n\right\};a_1=2;a_{n+1}=1-\frac{1}{a_n}$ for  $n\ge 1$. Let $P_n$  be the product of its first ‘n’ terms; then the value of $P_{2023}$  is _________

Since  $a_{n+2}=1-\frac{1}{a_{n+1}}=1-\frac{a_n}{a_{n-1}}=\frac{-1}{a_{n-1}}$
It follows that  $a_{n+3}=1-\frac{1}{a_{n+2}}=1+a_{n-1}=a_n$
So $\left\{a_n\right\}$ is a periodic sequence with period ‘3’ since  
 $a_1=2,a_2=1-\frac{1}{2}=\frac{1}{2},a_3=1-2=-1$
 $a_1{a}_2{a}_3=-1$
$P_{2023}=a_1{\left(a_1{a}_2{a}_3\right)}^{674}=2$