LK-$\phi$ proof of $\exists y Pby$
First of all, let me recommend then you use parentheses. So $\Phi = \{P(a,f(a))\}$, and you're trying to derive $\exists y P(b,y)$ from this.

If it were $\exists y P(a,y)$ instead, the proof would be very simple. You start by using the $\Phi$-rule to deduce $\vdash P(a,f(a))$, and then using the $\exists R$ rule you deduce $\vdash \exists y P(a,y)$.

As it stands, you cannot deduce $\exists y P(b,y)$ from $\Phi$ since $\exists y P(b,y)$ is not logically entailed from $\Phi$. For example, consider the interpretation in which $f$ is the identity and $P(x,y)$ is true only if $x=y=a$. Then $\Phi$ holds but $\exists y P(b,y)$ is false. Since LK-$\Phi$ is sound, you cannot deduce $\exists y P(b,y)$ from $\Phi$.