 Solution

Define $a_y=\min \{1,{\lambda }^y/7\}$. Given $Y=y$, the direct-return branch has probability $a_y$, while the fallback branch distributes its remaining mass uniformly over $0,\dots ,y$. The law of total probability therefore gives, for integer $x\ge 0$,The summation includes $y=x$: a failed direct-return test can still return $x$ through the discrete uniform distribution draw. There is no rejection loop and no additional normalization. Summing the displayed probability mass function over $x$ gives ${\sum }_y{p}_y[a_y+(1-a_y)(y+1)/(y+1)]=1$. At $\lambda =1$, for example, $a_y=1/7$ at every $y$; it does not become one. This is an instance of the accept-or-fallback mixture identity.