The lanthanoid that does NOT show +4 oxidation state is:

To determine which lanthanoid does not exhibit the $$+4$$ oxidation state, we examine the electronic configurations of the given elements. Although the most common oxidation state of lanthanoids is $$+3$$, certain elements can also exhibit $$+2$$ or $$+4$$ oxidation states if doing so results in a particularly stable electronic configuration, such as an empty $$f^0$$, half-filled $$f^7$$, or completely filled $$f^{14}$$ subshell.

Cerium has the electronic configuration

$$[Xe],4f^1,5d^1,6s^2.$$

On losing four electrons, it forms

$$Ce^{4+} : [Xe],4f^0.$$

Since this corresponds to an empty $$4f$$ subshell and a noble gas configuration, the $$+4$$ oxidation state is highly stable and commonly observed for cerium.

Terbium has the electronic configuration

$$[Xe],4f^9,6s^2.$$

Formation of $$Tb^{4+}$$ gives

$$[Xe],4f^7,$$

which is a half-filled $$4f$$ subshell. The exceptional stability associated with the $$4f^7$$ configuration allows terbium to readily exhibit the $$+4$$ oxidation state.

Dysprosium has the electronic configuration

$$[Xe],4f^{10},6s^2.$$

Although uncommon, dysprosium can attain the $$+4$$ oxidation state in certain compounds, producing

$$Dy^{4+} : [Xe],4f^8.$$

Such species have been reported in some solid-state fluoride complexes.

Europium has the electronic configuration

$$[Xe],4f^7,6s^2.$$

Loss of the two $$6s$$ electrons produces

$$Eu^{2+} : [Xe],4f^7,$$

which possesses the exceptionally stable half-filled $$4f^7$$ configuration. While europium can also form the usual $$+3$$ oxidation state, formation of $$Eu^{4+}$$ would require removing an electron from this highly stabilized half-filled $$4f$$ subshell. The energy required for this process is too large, making the $$+4$$ oxidation state essentially inaccessible under normal chemical conditions.

Therefore, among the given lanthanoids, Europium does not exhibit the $$+4$$ oxidation state.

Hence, the correct answer is

$$\boxed{\text{(C) Eu}}.$$