PromiseP is contained in PromiseBPP and PromiseBPP in BQP

If $\mathrm{\Pi }\in \mathrm{P}\mathrm{r}\mathrm{o}\mathrm{m}\mathrm{i}\mathrm{s}\mathrm{e}\mathrm{P}$, run its deterministic decider as a PTM that ignores all random bits. Its acceptance probability is $1$ on $Y$ and $0$ on $N$, so it satisfies [F2]. Thus $\mathrm{P}\mathrm{r}\mathrm{o}\mathrm{m}\mathrm{i}\mathrm{s}\mathrm{e}\mathrm{P}\subseteq \mathrm{P}\mathrm{r}\mathrm{o}\mathrm{m}\mathrm{i}\mathrm{s}\mathrm{e}\mathrm{B}\mathrm{P}\mathrm{P}$.

Now let $\mathrm{\Pi }\in \mathrm{P}\mathrm{r}\mathrm{o}\mathrm{m}\mathrm{i}\mathrm{s}\mathrm{e}\mathrm{B}\mathrm{P}\mathrm{P}$ be witnessed by a PTM $M$. Choose a strictly increasing positive integer polynomial $R(n)$ that bounds both its run time and the number of fair bits it can use on length-$n$ inputs. By [F3], using a uniform $R(n)$-bit string and ignoring its unused suffix preserves every acceptance probability.

For a binary word $w$ of length $m$, define one deterministic polynomial-time machine $M^{\mathrm{♯}}$ as follows. Search for an $n$ with $m=n+R(n)$; strict increase makes it unique if it exists. If none exists, output $0$. If it exists, split $w=x\mathrm{\parallel }r$ with $\mathrm{\mid }x\mathrm{\mid }=n$ and $\mathrm{\mid }r\mathrm{\mid }=R(n)$, simulate $M$ on $x$ using $r$ as its fair random tape, and output its accept bit. The search and simulation take polynomial time in $m$, since $n\le m$ and $R$ is a fixed polynomial. Whenever a fixed multitape stage in this construction or its later composition runs for at most $P(m)$ steps, a one-tape simulator stores its finitely many tape contents in aligned tracks of a finite product alphabet, with one head marker per tape. Each track uses at most $m+P(m)+O(1)$ cells, and a constant number of sweeps simulates each step in $O(m+P(m))$ time. Put the binary output on a designated track starting at cell $0$; a final cleanup sweep replaces each product symbol by its output-track component, leaving the exact output word followed by blanks on the final tape. Thus the construction has a one-tape polynomial-time implementation whenever that convention is needed.

Apply [F4] to $M^{\mathrm{♯}}$, then [F5] to emit the Boolean gate list at length $m(n)=n+R(n)$. Compose this generator with the polynomial-time map $n\mapsto m(n)$. Apply [F6] to that list. In the resulting QC14 family on input $x$, designate the next $R(n)$ auxiliary wires as the random register, apply $H$ to each from its initial zero state, and then run the reversible compiler on the $x$ and random registers with a zero target and clean work. Starting from the compiled list with $g$ gates, prefix Hadamard records on wire indices $n,\dots ,n+R(n)-1$, change the unary input-wire count to $n$, and rewrite the self-delimiting gate count to $g+R(n)$; retain the same total-wire count, target tuple and compiled gate records. The fixed composition of these list operations runs in polynomial time in $n$ because all intermediate Boolean and QC14 descriptions have polynomial length. The one-tape simulation described in step 2.1 therefore yields one fixed deterministic one-tape generator that leaves the complete QC14 description as its final tape word, as required for the full-list family.

By [F8]–[F9], the Hadamards prepare $2^{-R(n)\mathrm{/}2}{\sum }_{r\in \{0,1{\}}^{R(n)}}\mathrm{\mid }r⟩$ on the zero-initialized random register. For each basis $r$, the reversible circuit writes exactly the bit $M(x;r)$ into the target and returns its work to zero. Distinct $r$ values remain in orthogonal computational-basis states of the unmeasured random register. The one-tape generator from step 3.1 satisfies the full-list uniformity condition in [F7]. Therefore, by [F10], the target's acceptance probability is $2^{-R(n)}\mathrm{\#}\{r\in \{0,1{\}}^{R(n)}:M(x;r)\text{ accepts}\},$ which is exactly the PTM acceptance probability in [F3]. Its YES and NO bounds in [F2] are the BQP bounds in [F7]. Hence $\mathrm{P}\mathrm{r}\mathrm{o}\mathrm{m}\mathrm{i}\mathrm{s}\mathrm{e}\mathrm{B}\mathrm{P}\mathrm{P}\subseteq \mathrm{B}\mathrm{Q}\mathrm{P}$.

For an ordinary language $L\in \mathrm{B}\mathrm{P}\mathrm{P}$, use the total promise pair $(L,\{0,1{\}}^{\ast }\setminus L)$. Its BPP PTM satisfies the PromiseBPP definition [F2], so step 4.1 gives a BQP family for the language. Thus $\mathrm{B}\mathrm{P}\mathrm{P}\subseteq \mathrm{B}\mathrm{Q}\mathrm{P}$.