Is following SAT case in $\mathsf{P}$?

Question

f[1](2, 3, 20) : 10011010
f[2](4, 20, 21) : 10101001
f[3](4, 20, 22) : 10010110
f[4](2, 4, 23) : 10011010
f[5](5, 23, 24) : 10101001
f[6](5, 21, 23) : 01101001
f[7](2, 5, 26) : 10011010
f[8](2, 3, 33) : 10101001
f[9](3, 22, 33) : 10000110
f[10](3, 21, 40) : 10100110
f[11](3, 21, 41) : 01101001
f[12](21, 41, 42) : 10101001
f[13](40, 42, 44) : 10010110
f[14](3, 4, 45) : 10101001
f[15](26, 45, 46) : 10101001
f[16](26, 45, 47) : 10010110
f[17](24, 46, 47) : 11001001
f[18](24, 44, 47) : 01101001

Each f[i](a, b, c) : truth-table means some boolean formula of 3 variables $x_a, x_b, x_c$. The main formula is conjunction of them all (I just converted CNF for 53 factorization circuit to such form and managed to reduce it to given instance).

$$\Phi = f_1 \land f_2 \land\ ...\ \land f_n$$

What I think is that now I need only logarithmic (or even constant) amount of all possible assignments to see if it's satisfiable.

For example, I can take $x_2 = 1$ and it'll automatically result in $x_{20} = 0, x_{21} = 0, x_4 \neq x_{22}, x_{23} = 0,$ etc. It seems that half of all variables will automatically be assigned, if I'll just put one.

So, is it in $\mathsf{P}$?