NP is contained in P with a Sharp-P oracle

Enlarge $p$ if needed to a nonnegative integer-valued polynomial bound, and put $m=p(\mathrm{\mid }x\mathrm{\mid })$. Define $h(x)=\mathrm{\mid }\{u\in \{0,1{\}}^{\ast }:\mathrm{\mid }u\mathrm{\mid }\le m,R(x,u)=1\}\mathrm{\mid }$. A binary-branching polynomial-time machine guesses a word $y$ of length $m+1$. It rejects if $y$ has no zero; otherwise it writes $y=1^{m-\mathrm{\mid }u\mathrm{\mid }}0u$ using the first zero, and accepts exactly when $R(x,u)=1$. This encoding is a bijection between words $u$ of length at most $m$ and guessed words $y$ with a zero. Thus the accepting-path count is exactly $h(x)$, so $h\in \mathrm{\#}\mathrm{P}$.

By [L3], compute in polynomial time a formula $r(x)$ with $h(x)=\mathrm{N}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{S}\mathrm{A}\mathrm{T}(r(x))$. A deterministic oracle machine queries $r(x)$ and accepts iff the returned binary integer is nonzero. Since $h(x)\le 2^{m+1}-1$, the answer has at most $m+1$ bits. By the verifier definition, $h(x)>0$ exactly when $x\in L$.

This oracle machine decides $L$, proving the containment.