On the prime number theorem in arithmetic progression The prime number theorem tells us that , if $\pi\left(x\right)$ denotes the number of primes less than or equal to $x$, we have $$\pi\left(x\right)\sim\frac{x}{\log x}.$$
In a similar manner considered $1\leq a \leq q$ with $(a,q)=1$ and defined $\pi\left(x,a,q\right)$ the number of primes less than or equal to $x$ congruous $a\,\textrm{mod}\, q$ and $\phi\left(n\right)$ the number of minor numbers and coprime with $n$, we have $$\pi(x,a,q)\thicksim\frac{1}{\phi(q)}\frac{x}{\log x}.$$
If $q$ is "small" you have asymptotic formulas for $\pi\left(x,a,q\right)$ (see the Siegel - Walfisz theorem). For any $q$ we have the estimate
$$\pi(x,a,q)\gg\frac{1}{\phi(q)}\frac{x}{\log x}.$$
I would like to know if there is an estimate of the type $$\pi(x,a,q)\ll\frac{1}{\phi(q)}\frac{x}{\log x}$$ for any $q$.
I hope I was clear! Sorry for my bad english!
 A: Siegel - Walfisz theorem states 
$$\psi(x,q;a)=x/\phi(q)+O(x/\log^Ax)$$when q is small;q is big ,the results is trival .  (When is the Siegel-Walfisz theorem non-trivial?)
if $q \ll log^Ax$,we have
$$\sum_{x\equiv 1 \bmod q } \Lambda(n)\ge  (1-\epsilon) x/\varphi(q).$$
(Prime numbers in arithmetic progressions : uniformity with respect to the modulus
)
I think q is big, there is no the $\ll$ results as you asked. Since when q is big ,there may be existing Siegel-zeros, then there exist constants 0 < β−, β+ < 1 such that
$x^{\beta_{-}}/\beta_{-}\ll x-\phi(q)\pi(x,q;a) \log x \ll x^{\beta}_{+}/ \beta _{+}$.
(see section 3 , and http://www.dms.umontreal.ca/~andrew/PDF/ItalySurvey.pdf)
or see 
Theorem (Siegel-Walfisz with a twist). in (Chebyshev function in arithmetic progressions)
A: For $x\leq\phi(q)$ the estimate $\pi(x,a,q)\ll\frac{1}{\phi(q)}\frac{x}{\log x}$ would imply $\pi(x,a,q)\ll\frac{1}{\log x}$, i.e. $\pi(x,a,q)=0$ for large $x$ which is clearly false. So a bound you envision can only hold for $x$ slightly above $\phi(q)$. On the other hand, for any $\epsilon>0$, the Brun-Titchmarsh inequality implies
$$\pi(x,a,q)\ll_\epsilon\frac{1}{\phi(q)}\frac{x}{\log x},\qquad x>q^{1+\epsilon}.$$
