Bond polarity and ionic character go hand in hand: We can say that polarity and ionic character are positively correlated. The classical way to determine either is to compare the electronegativity difference (ΔEN) between the bonding atoms. The greater the difference in electronegativity, the larger the dipole moment, the greater the polarity, and the greater the ionic character of the bond.
The Pauling electronegativity scale, which should be provided in most chemistry texts and is certainly available online, can be used to determine ΔEN. As a very general rule of thumb, an ΔEN that is equal to 2.0 or greater indicates an ionic bond. We don't have to worry about that here, though, since we're only interested in polarity. But this is just some conceptual relation to consider; indeed, ionic bonds can be seen as very polar covalent bonds (or covalent bonds with significant ionic character).
There is no known bond that has a 100% ionic character (ΔEN = 4), i.e., a purely ionic bond. But there are many bonds that we know of with 0% ionic character (ΔEN = 0), i.e., a completely nonpolar bond. We can think about this logically without even knowing specific electronegativity values: If the difference in electronegativity between two bonding atoms is 0, then the two atoms must have equivalent electronegativities. And when can this be the case? Well, atoms from the same element have the same electronegativity. So one case of completely nonpolar bonding would be covalent bonding between atoms of the same element.
In the question given, there are two such cases: P4 and Br-Br. P4 is a tetrahedral molecule composed of four covalently bonded P atoms, each one bearing a lone electron pair and bonded to the other three. With each P having the same electronegativity (ΔEN = 0), no P-P bond in P4 has a dipole moment, and the P4 molecule itself is nonpolar. Br-Br is how bromine is found naturally (as diatomic Br₂); since this is a covalent bond between the atoms of the same element, the ΔEN = 0 and there is no dipole moment. A bromine molecule is completely nonpolar.
Fluorine (F) has the greatest electronegativity among the elements, so whether the Te-F or the Cl-F bond has the greater polarity (i.e., the greater ΔEN) really depends on the electronegativity of Te and Cl.
But let's think about this first. Both Cl and F are halogens; Cl is in the period right below F, and, following periodic trends, Cl itself is quite electronegative. Generally, nonmetals don't form ionic bonds with each other (the underlying reason is complicated and not relevant here, but that notion agrees with ΔEN between any two nonmetals being less than 2.0). Nonmetallic elements can form very polar covalent bonds whose ΔEN is very close to 2.0, however.
Tellurium (Te) is a metalloid that is three periods below F. Electronegativity generally decreases as you go down a group. Even though Te and Cl are not in the same group, the fact that Te is two periods below should give a strong presumption that Te has a lower electronegativity than Cl. Moreover, electronegativity increases you move left to right on the periodic table (that is, as the group number increases); Te is in a group to the left of fluorine's and chlorine's. So, using periodic trends alone, we know that Te is in a group to the left of and below that of Cl, both of which trend toward decreasing electronegativity. Thus, it would be reasonable to assume that Te would have a smaller electronegativity than Cl. And so, the ΔEN of Te-F > ΔEN of Cl-F, which means the Te-F bond should have the greater polarity. And, indeed, this is true using the Pauling scale: The EN of F is 4.0; the EN of Cl is 3.0; and the EN of Te is 2.1.
In all, given that the ΔEN of P-P = 0; ΔEN of Br-Br = 0; ΔEN of Te-F = 1.9; and ΔEN of Cl-F = 1.0; the Te-F bond has the greatest polarity.
