Boundedness $L^2$-orthogonal projection Let $\varphi \in L^\infty ([0,1]^d)$ and $P:L^\infty ([0,1]^d)\to \mathbb Q_p ([0,1]^d)$ be defined as
$$
\int_{[0,1]^d} P (\varphi) q =   \int_{[0,1]^d}\varphi q\qquad\forall\ q\in  \mathbb Q_p ([0,1]^d).
$$
Is it true that there exists $C>0$, independent of $\varphi$, such that $\| P(\varphi) \|_{_{L^{\infty}} ([0,1]^d)} \le C \| \varphi \|_{L^{\infty}  ([0,1]^d)}$?
Here, $\mathbb Q_p ([0,1]^d)$ denotes the space of polynomials of degree at most $p$ in every coordinate direction.

[Incomplete answer]
For $d=1$ we can prove that, for a given $p$, there is a $C=C_p$ for which the projection operator in bounded by $C_p$.
For a given $p$, let $\varphi\in L^\infty([0,1])$, then $P(\varphi)(x)=\sum_{k=0}^p v_kx^k$ must satisfy a linear system of the form $Av=b$, where $A$ is the Hilbert matrix, and
$$
b_i = \int_0^1 \varphi(x)x^idx, i=0,...,p
$$
This matrix is always invertible, thus we can get $v=A^{-1}b$. Hence, all $v_i$ will be of the form
$$
v_i =\int_0^1 \varphi(x)p_i(x)dx
$$
where $p_i(x)\in\mathbb{Q}_p$ are polynomials whose coefficients are the values in the i-row of $A^{-1}$. Hence
$$
|v_i| = \left|\int_0^1 \varphi(x)p_i(x)dx\right|\\
\le \|\varphi\|_{L^\infty}\int_0^1 |p_i(x)|dx\\
= C_{i,p}\|\varphi\|_{L^\infty}
$$
Hence for all $x$, $|P(\varphi)|(x)=|\sum_{k=0}^p v_kx^k|\le \sum_{k=0}^p |v_k|x^k\le \|\varphi\|_{L^\infty}(C_{0,p} + ... + C_{p,p}) $

[Guess]
For $d>1$, a strategy to study this is to attempt to generalize the argument used for $d=1$. Namely, let $\varphi\in L^\infty([0,1]^d)$, then $P(\varphi)(x_1, ..., x_d)=\sum_{k_1,...,k_d=0}^p v_{k_1,...,k_d} x_1^{k_1}...x_d^{k_d}$ must satisfy, for all $j_1,...,j_d$,
$$
\int_{[0,1]^d} \varphi(x_1, ..., x_d)x_1^{j_1}...x_d^{j_d} dx_1...dx_d = \\
\int_{[0,1]^d} \left(\sum_{k_1,...,k_d=0}^p v_{k_1,...,k_d}x_1^{k_1}...x_d^{k_d}\right)x_1^{j_1}...x_d^{j_d} dx_1...dx_d
$$
which yields a linear system of the form $\bar{A}\bar{v}=\bar{b}$. However, in this case I am not sure such $\bar{A}$ would be invertible. In case it were, though, we can prove the result using similar arguments to those for the case $d=1$
