i.e., $P=q \times d$

here compound is $HCl$,you know as well, $HCl$ is ionic compound in which one electron transfers from hydrogen to chlorine.

so, magnitude of charge on dipole, $q =1.6 \times 10^{-19} \,C$

and seperation between charges $=$ bond length $=1.26 A ^{\circ}=1.26 \times 10^{-10} \,m$

so, dipole moment, $P =1.6 \times 10^{-19} \times 1.26 \times 10^{-10}\, Cm$ $=2.016 \times 10^{-29}\, Cm$

we know, $1 D =3.335 \times 10^{-30}\, Cm$ [ Debay, $D$ is the unit of dipole moment ] $=2.016 \times 10^{-29} /\left(3.335 \times 10^{-30}\right)$

$=20.16 / 3.335$

$=6.0449\, D$

now, percentage ionic character $=$ experimental value/ theoretical value $\times 100$ $=1.03 \,D / 6.0449 \,D \times 100$

$=0.170 \times 100$

$=17.00\, \%$